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Page No 30.25:

Question 1:

(i) 2x

(ii) 1x

(iii) 1x3

(iv) x2+1x

(v) x2-1x

(vi) x+1x+2

(vii) x+23x+5

(viii) k xn

(ix) 13-x

(x) x2 + x + 3

(xi) (x + 2)3

(xii) x3 + 4x2 + 3x + 2

(xiii) (x2 + 1) (x − 5)

(xiv) 2x2+1

(xv) 2x+3x-2

Answer:

i ddxf(x)=limh0fx+h-fxh                 =limh02x+h-2xh                 =limh02x-2x-2hhx(x+h)                 =limh0-2hhx(x+h)                 =limh0-2x(x+h)                 =-2x2
ii ddxf(x)=limh0fx+h-fxh                 =limh01x+h-1xh                 =limh0x-x+hhxx+h×x+x+hx+x+h                 =limh0x-x-hhxx+hx+x+h                 =limh0-hhxx+hx+x+h                 =limh0-1xx+hx+x+h                 =-1xxx+x                 =-1x× 2x                 =-12x32                 =-12x-32
iii ddxf(x)=limh0fx+h-fxh                  =limh01(x+h)3-1x3h                  =limh0x3-(x+h)3h(x+h)3x3                  =limh0x3-x3-3x2h-3xh2-h3h(x+h)3x3                  =limh0-3x2h-3xh2-h3h(x+h)3x3                  =limh0h-3x2-3xh-h2h(x+h)3x3                  =limh0-3x2-3xh-h2(x+h)3x3                  =-3x2x6                  =-3x4                  =-3x-4
iv ddxf(x)=limh0fx+h-fxh                     =limh0(x+h)2+1x+h-x2+1xh                     =limh0x2+2xh+h2+1x+h-x2+1xh                     =limh0x3+2x2h+h2x+x-x3-x2h-x-hxh(x+h)                     =limh0x2h+h2x-hx(x+h)                     =limh0h(x2+hx-1)xh(x+h)                     =limh0x2+hx-1x(x+h)                     =x2-1x2

v ddxf(x)=limh0fx+h-fxh                     =limh0(x+h)2-1x+h-x2-1xh                     =limh0x2+2xh+h2-1x+h-x2-1xh                     =limh0x3+2x2h+h2x-x-x3-x2h+x+hxh(x+h)                     =limh0x2h+h2x+hx(x+h)                     =limh0h(x2+hx+1)xh(x+h)                     =limh0x2+hx+1x(x+h)                     =x2+1x2

vi ddxf(x)=limh0fx+h-fxh                     =limh0x+h+1x+h+2-x+1x+2h                     =limh0x+h+1x+2-x+h+2x+1hx+h+2x+2                     =limh0x2+2x+hx+2h+x+2-x2-x-hx-h-2x-2hx+h+2x+2                     =limh0hhx+h+2x+2                     =limh01x+h+2x+2                     =1x+0+2x+2                     =1x+22

vii ddxf(x)=limh0fx+h-fxh                     =limh0x+h+23x+h+5-x+23x+5h                     =limh0x+h+23x+3h+5-x+23x+5h                     =limh0x+h+23x+5-3x+3h+5x+2h3x+3h+53x+5                     =limh03x2+3xh+6x+5x+5h+10-3x2-3xh-5x-6x-6h-10h3x+3h+53x+5                     =limh0-hh3x+3h+53x+5                     =limh0-13x+3h+53x+5                     =-13x+52

viii ddxf(x)=limh0fx+h-fxh                     =limh0 k x+hn-kxnh                     =limx+h-x0 k x+hn-xnx+h-xHere, we have: limxaxm-amx-a=m am-1                     =k n xn-1

ix ddxf(x)=limh0fx+h-fxh                     =limh0 13-x-h-13-xh                     =limh0 3-x-3-x-hh3-x3-x-h                     =limh0 3-x-3-x-hh3-x3-x-h× 3-x+3-x-h3-x+3-x-h                     =limh0 3-x-3+x+hh3-x3-x-h3-x+3-x-h                     =limh0 hh3-x3-x-h3-x+3-x-h                     =limh0 13-x3-x-h3-x+3-x-h                     =13-x3-x-03-x+3-x-0                     =13-x 23-x                     =123-x32

x ddxf(x)=limh0fx+h-fxh                     =limh0 x+h2+x+h+3-x2+x+3h                     =limh0x2+h2+2xh+x+h+3-x2-x-3h                     =limh0h2+2xh+hh                     =limh0 h(h+2x+1)h                     =limh0 h+2x+1                     =0+2x+1                     =2x+1

xi ddxf(x)=limh0fx+h-fxh                     =limh0 x+h+23-x+23h                     =limh0x+h+2-x-2x+h+22+x+h+2x+2+x+22h                     =limh0hx+h+22+x+h+2x+2+x+22h                     =limh0 x+h+22+x+h+2x+2+x+22                     =x+0+22+x+0+2x+2+x+22                     =x+22+x+22+x+22                     =3x+22

xii ddxf(x)=limh0fx+h-fxh                     =limh0 x+h3+4x+h2+3x+h+2-x3+4x2+3x+2h                     =limh0x3+3x2h+3xh2+h3+4x2+4h2+8xh+3x+3h+2-x3-4x2-3x-2h                     =limh0 3x2h+3xh2+h3+4h2+8xh+3h+2h                     =limh0 h3x2+3xh+h2+4h+8x+3h                     =limh03x2+3xh+h2+4h+8x+3                     =3x2+8x+3

xiii x2+1x-5=x3-5x2+x-5ddxf(x)=limh0fx+h-fxh                     =limh0 x+h3-5x+h2+x+h-5-x3-5x2+x-5h                     =limh0 x3+3x2h+3xh2+h3-5x2-5h2-10xh+x+h-5-x3+5x2-x+5h                     =limh0 3x2h+3xh2+h3-5h2-10xh+hh                     =limh0 h 3x2+3xh+h2-5h-10x+1h                     =limh0 3x2+3xh+h2-5h-10x+1                     =3x2-10x+1

xiv ddxf(x)=limh0fx+h-fxh                     =limh0 2x+h2+1-2x2+1h                     =limh02x2+2h2+4xh+1-2x2+1h                     =limh02x2+2h2+4xh+1-2x2+1h×2x2+2h2+4xh+1+2x2+12x2+2h2+4xh+1+2x2+1                     =limh0 2x2+2h2+4xh+1-2x2-1h2x2+2h2+4xh+1+2x2+1                     =limh0 h2h+4xh2x2+2h2+4xh+1+2x2+1                     =limh0 2h+4x2x2+2h2+4xh+1+2x2+1                     =4x2x2+1+2x2+1                     =4x22x2+1                     =2x2x2+1

xv ddxf(x)=limh0fx+h-fxh                     =limh0 2x+h+3x+h-2-2x+3x-2h                     =limh0 2x+2h+3x-2-x+h-22x+3hx+h-2x-2                     =limh0 2x2+2xh+3x-4x-4h-6-2x2-2xh+4x-3x-3h+6hx+h-2x-2                     =limh0 -7hhx+h-2x-2                     =limh0 -7x+h-2x-2                     =-7x-2x-2                     =-7x-22

Page No 30.25:

Question 2:

Differentiate each of the following from first principles:

(i) ex

(ii) e3x

(iii) eax + b

(iv) x ex

(v) − x

(vi) (−x)−1

(vii) sin (x + 1)

(viii) cosx-π8

(ix) x sin x

(x) x cos x

(xi) sin (2x − 3)

Answer:

i ddxf(x)=limh0fx+h-fxhddxex=limh0e-(x+h)-e-xh             =limh0e-xe-h-e-xh             =limh0e-xe-h-1h             =-e-xlimh0e-h-1-h             =-e-x1             =-e-x

ii ddxf(x)=limh0fx+h-fxhddxe3x=limh0e3(x+h)-e3xh             =limh0e3xe3h-e3xh             =limh0e3xe3h-13h             =3 e3xlimh0e3h-13h             =3 e3x1             =3 e3x

iii ddxf(x)=limh0fx+h-fxhddxeax+b=limh0ea(x+h)+b-eax+bh             =limh0eax+beah-eax+bh             =limh0eax+beah-1h             =a eax+blimh0eah-1ah             =a eax+b1             =a eax+b

iv ddxf(x)=limh0fx+h-fxhddxx ex=limh0(x+h )e(x+h)-x exh                =limh0(x+h) exeh-x exh                =limh0x exeh+hexeh-x exh                =limh0x exeh-x exh+limh0h exehh                =limh0x exeh-1h+limh0exeh                =xex1+exe0                =xex+ex

v ddxfx=limh0fx+h-fxhddx-x=limh0-x+h--xh              =limh0-x-h+xh              =limh0-hh              =limh0-1              =-1

vi -x-1=1-x ddxfx=limh0fx+h-fxhddx1-x =limh01-x+h-1-x  h                =limh0-1x+h+1xh                =limh0-x+x+hh x x+h                =limh0hh x x+h                =limh01 x x+h                =1x.x                =1x2

vii ddxfx=limh0fx+h-fxhddxsin x+1=limh0sin x+h+1-sin x+1 hWe know:sin C-sin D=2 cos C+D2 sin C-D2                       =limh02 cos x+h+1+x+12 sin x+h+1-x-12h                       =limh02 cos 2x+h+22 sin h2 h                       =2limh0 cos 2x+h+22   limh0 sin h2 h2×12                       =2 cos x+1 ×12                       =cos x+1

viii ddxfx=limh0fx+h-fxhddxcos x-π8=limh0cos x+h-π8-cos x-π8hWe know:cos C-cos D=-2 sin C+D2 sin C-D2                        =limh0-2 sin x+h-π8+x-π82 sin x+h-π8-x+π82h                        =limh0-2 sin 2x+h-π42 sin h2 h                        =-2limh0 sin 2x+h-π42 limh0 sin h2 h2×12                        =-2 sin x-π8 ×12                        =-sin x-π8

ix ddxf(x)=limh0fx+h-fxh                  =limh0 x+h sinx+h - x sin xh                  =limh0x+hsin x cos h + cos x sin h- x sin xh                  =limh0 x sin x cos h + x cos x sin h +h sin x cos h + h cos x sin hh                  =limh0 x sin x cos h  - x sin x + x cos x sin h +h sin x cos h + h cos x sin h - x sin x h                  =x sin x limh0   cos h -1h+x cos x limh0 sin h h+sin xlimh0 cos h+cos xlimh0 sin h                  =x sin x limh0 -2 sin2 h2h24×h4+ x cos x 1+ sin x 1+ cos x 0                 = x sin x × -h2 + x cos x 1+ sin x 1+ cos x 0                  =-2x sin x 120+ x cos x + sin x                   = x cos x + sin x  

x ddxf(x)=limh0fx+h-fxh                   =limh0 x+h cos x+h - x cos xh                   =limh0x+hcos x cos h-sin x sin h- x cos xh                   =limh0 x cos x cos h - x sin x sin h+h cos x cos h - h sin x sin h- x cos xh                   =limh0 x cos x cos h  - x cos x - x sin x sin h+h cos x cos h - h sin x sin hh                   =x cos x limh0   cos h -1h-x sin x limh0 sin h h+cos xlimh0 cos h+sin xlimh0 sin h                   =x cos x limh0 -2 sin2 h2h24×h4-x sin x 1+cos x 1+ sin  x 0                  =xcosx  lim h0-h2-x sin x 1+cos x 1+ sin  x 0                  =-x cos x 0-x sin x+ cos x                    = -x sin x+ cos x  

xi ddxf(x)=limh0fx+h-fxh                    =limh0 sin 2x+2h-3-sin 2x-3hWe know:sin C-sin D=2 cos C+D2 sinC-D2                     =limh0 2 cos 2x+2h-3+2x-32 sin 2x+2h-3+2x-32 h                    =limh0 2 cos 4x+2h-62 sin h h                    =limh0 2 cos 4x+2h-62 limh0 sin hh                    =2 cos 4x-62 1                    = 2 cos 2x-3 



Page No 30.26:

Question 3:

Differentiate each of the following from first principles:

(i) sin 2x

(ii) sin xx

(iii) cos xx

(iv) x2 sin x

(v) sin (3x+1)

(vi) sin x + cos x

(vii) x2 ex

(viii) ex2+1

(ix) e2x

(x) eax+b

(xi) ax

(x) 3x2

Answer:

i ddxf(x)=limh0fx+h-fxh                   =limh0 sin 2x+2h-sin 2xh×sin 2x+2h+sin 2xsin 2x+2h+sin 2x                   =limh0sin 2x+2h- sin 2xh sin 2x+2h+sin 2xWe have:sin C-sin D= 2 cos C+D2 sin C-D2                   =limh02 cos 2x+2h+2x2 sin 2x+2h-2x2h sin 2x+2h+sin 2x                   =limh02 cos 2x+h sin hh sin 2x+2h+sin 2x                   =limh0 2 cos 2x+h   limh0 sin h hlimh01sin 2x+2h+sin 2x                    = 2 cos 2x 1 1sin 2x+sin 2x                   =2 cos 2x2sin 2x                   = cos 2xsin 2x

ii ddxf(x)=limh0fx+h-fxh                    =limh0 sin x+hx+h-sin xxh                    =limh0 x sin x+h- x+h sin x h x x+h                    =limh0 x sin x cos h + cos x sin h- x sin x - h sin xh x x+h                    =limh0 x sin x cos h +x cos x sin h- x sin x - h sin xh x x+h                    =limh0 x sin x cos h- x sin x +x cos x sin h - h sin xh x x+h                    =x sin xlimh0cos h -1h+x cos x x limh0 sin hhlimh0 1x+h-sin xxlimh01x+h                     =x sin x limh0 -2 sin2 h2h+x cos x x limh0 sin hhlimh0 1x+h-sin xxlimh01x+h                     =x sin x limh0 -2 sin2 h2h24×h4+x cos x x limh0 sin hhlimh0 1x+h-sin xxlimh01x+h                    =-x sin x ×limh0h2+x cos x x limh0 sin hhlimh0 1x+h-sin xxlimh01x+h                    =-x sin x 12 0+cos xx-sin xx2                    =cos xx-sin xx2                    =x cos x-sin xx2 

iii ddxf(x)=limh0fx+h-fxh                   =limh0 cos x+hx+h-cos xxh                   =limh0 x cos x+h- x+h cos x h x x+h                   =limh0 x cos x cos h - sin x sin h- x cos x - h cos xh x x+h                   =limh0 x cos x cos h -x sin x sin h- x cos x - h cos xh x x+h                   =limh0 x cos x cos h - x cos x-x sin x sin h - h cos xh x x+h                   =xcos xlimh0cos h -1h-xsin x x limh0 sin hhlimh0 1x+h-cos xxlimh01x+h                    =x cos x limh0 -2 sin2 h2h24×h4-xsin x x limh0 sin hhlimh0 1x+h-cos xxlimh01x+h                  limh0 sin2 h2h24=limh0sin h2h2×limh0sin h2h2=1×1, i.e. 1                    = -x cosxlimh0 h2-xsin x x limh0 sin hhlimh0 1x+h-cos xxlimh01x+h                  =-x cos x ×0-sin x 11x-cos x x1x                   =0-sin xx-cos xx2                   =-sin xx-cos xx2                   =-x sin x-cos xx2 

iv ddxf(x)=limh0fx+h-fxh                      =limh0 x+h2 sin x+h-x2 sin xh                      =limh0 x2+h2+2xhsin x cos h + cos x sin h-x2 sin xh                      =limh0 x2 sin x cos h + x2 cos x sin h + h2 sin x cos h +h2 cos x sin h+2xh sin x cos h +2xh cos x sin h-x2 sin xh                      =limh0x2 sin x cos h-x2 sin x + x2 cos x sin h + h2 sin x cos h +h2 cos x sin h+2xh sin x cos h +2xh cos x sin hh                      =x2 sin xlimh0cos h -1h+x2 cos xlimh0sin hh+sin x limh0 h cos h+ cos x limh0 h sin h +2x sin xlimh0 cosh +2x cos x limh0 sin h                      =x2 sin x limh0 -2 sin2 h2h24×h4+x2 cos xlimh0sin hh+sin x limh0 h cos h+ cos x limh0 h sin h +2x sin xlimh0 cosh +2x cos x limh0 sin h          limh0 sin2 h2h24=limh0sin h2h2×limh0sin h2h2=1×1, i.e. 1                      =-x2sinx× limh0 h2+x2 cos xlimh0sin hh+sin x limh0 h cos h+ cos x limh0 h sin h +2x sin xlimh0 cosh +2x cos x limh0 sin h                             =-x2 sin x ×0+x2 cos x 1+ sin x 0+ cos x 0+ 2x sin x 1+2x cos x 0                      =0+x2 cos x+ 2x sin x                      =0+x2 cos x+ 2x sin x                      =x2 cos x+ 2x sin x 

v ddxf(x)=limh0fx+h-fxh                        = limh0 sin 3x+h+1-sin 3x+1h                                         =limh0 sin 3x+3h+1-sin 3x+1h×sin 3x+3h+1+sin 3x+1sin 3x+3h+1+sin 3x+1                      =limh0sin 3x+3h+1-sin 3x+1h sin 3x+3h+1+sin 3x+1We have:  sin C-sin D= 2 cos C+D2 sin C-D2                      =limh02 cos 3x+3h+1+3x+12 sin 3x+3h+1-3x-12h sin 3x+3h+1+sin 3x+1                      =limh02 cos 6x+3h+22 sin 3h2h sin 3x+3h+1+sin 3x+1                      =limh0 2 cos 6x+3h+22   limh0 sin 3h 2h×32×32×limh01sin 3x+3h+1+sin 3x+1                       = 2 cos 3x+1 ×32 ×1sin 3x+1+sin 3x+1                      =3 cos  3x+12sin 3x+1 

vi ddxf(x)=limh0fx+h-fxh                    =limh0 sin x+h+cos x+h-sin x-cos xh                    =limh0sin x+h -sin xh+limh0 cos x+h-cos xhWe have: sin C-sin D= 2 cos C+D2 sin C-D2And, cos C-cos D=-2 sin C+D2 sin C-D2                    =limh0 2 cos 2x+h2 sin h2h+limh0 -2 sin 2x+h2 sin h2h                    =2 limh0 cos 2x+h2 limh0 sin h2h2×12 -2 limh0 sin 2x+h2  limh0 sin h2h2×12                    =2 cos x× 12-2 sin x ×12                    =cos x - sin x 

vii ddxf(x)=limh0fx+h-fxhddxx2 ex=limh0(x+h)2e(x+h)-x2exh                =limh0(x2+2xh+h2)exeh-x2exh                =limh0x2exeh+2xhexeh+h2exeh-x2exh                =limh0x2exeh-x2exh+limh02 x h exehh+limh0h2exehh                =limh0x2exeh-1h+limh02 x exeh+limh0 h exeh                =x2ex1+2xex1+0                =x2ex+2xex                =x2+2x ex

viii ddxf(x)=limh0fx+h-fxhddxex2+1=limh0e(x+h)2+1-ex2+1h                   =limh0ex2+h2+2xh+1-ex2+1h                   =limh0ex2+1eh2+2xh-ex2+1h                   =limh0ex2+1ehh+2x-1h×h+2xh+2x                   =ex2+1limh0 ehh+2x-1hh+2x limh0 h+2x                   =ex2+11 2x                   =2x ex2+1

ix ddxf(x)=limh0fx+h-fxhddxe2x=limh0e2(x+h)-e2xh                   =2 limh0e2x+2h-e2x2x+2h-2x                   =2 limh0 e2xe2x+2h-2x-12x+2h2-2x2                   =2 e2x limh0 e2x+2h-2x-12x+2h-2x2x+2h+2x                   =2 e2x limh0 e2x+2h-2x-12x+2h-2x limh012x+2h+2x                   =2 e2x 1122x                   =e2x2x

x ddxf(x)=limh0fx+h-fxhddxeax+b=limh0eax+ah+b-eax+bh                        =a limh0eax+ah+b-eax+bax+ah+b-ax+b                        =a limh0eax+beax+ah+b-ax+b-1ax+ah+b2-ax+b2                        =a eax+b limh0 eax+ah+b-ax+b-1ax+ah+b-ax+bax+ah+b+ax+b                        =a eax+b limh0 eax+ah+b-ax+b-1ax+ah+b-ax+b limh01ax+ah+b+ax+b                        =a eax+b 112ax+b                        =a eax+b2ax+b

xi ddxf(x)=limh0fx+h-fxhddxax=limh0ax+h-axh                 =limh0axax+h-x-1x+h-x                  =axlimh0ax+h-x-1x+h2-x2                 =axlimh0ax+h-x-1x+h-xx+h+x                 =axlimh0ax+h-x-1x+h-x limh01x+h+x                 =ax loge a 12x                 =12xax loge a

xii ddxf(x)=limh0fx+h-fxhddx3x2=limh03x+h2-3x2h              =limh0 3x2+2xh+h2-3x2h             =limh0 3x2 3x2+2xh+h2-x2-1h×h+2xh+2x              =3x2 limh0 3hh+2x-1hh+2xlimh0h+2x              =3x2 log 3 2x              =2x 3x2 log 3

Page No 30.26:

Question 4:

(i) tan2x

(ii) tan (2x + 1)

(iii) tan 2x

(iv) tan x

Answer:

i ddxf(x)=limh0fx+h-fxh                =limh0tan2 x+h-tan2 xh                =limh0 tan x+h+tan xtan x+h-tan xh                =limh0 sin x+hcos x+h+sin xcos xsin (x+h)cos (x+h)-sin xcos x h                =limh0 sin x+h cos x + cos x+h sin x sin x+h cos x - cos x+h sin x h cos2 x cos2 x+h                 =limh0 sin 2x+hsin hh cos2 x cos2 x+h                 =1cos2 xlimh0 sin 2x+h  limh0 sin hh limh0 1cos2x+h                =1cos2 x sin 2x 11cos2 x                =1cos2 x 2 sin x cos x 1cos2 x                =2× sin x cos x×1cos2 x                =2 tan x sec2x

ii ddxf(x)=limh0fx+h-fxh                   =limh0 tan 2x+2h+1-tan 2x+1h                   =limh0sin 2x+2h+1cos 2x+2h+1-sin 2x+1cos 2x+1h                   =limh0 sin 2x+2h+1 cos 2x+1-cos 2x+2h+1 sin 2x+1 h cos 2x+2h+1 cos 2x+1                   =limh0 sin 2x+2h+1-2x-1h cos 2x+2h+1 cos 2x+1                   =1cos 2x+1limh0 sin 2h2h×2 limh01 cos 2x+2h+1                   =1cos 2x+1 ×2×1cos 2x+1                   =2cos22x+1                   =2 sec2 2x+1

iii ddxf(x)=limh0fx+h-fxh                    =limh0 tan 2x+2h-tan 2xh                   =limh0sin 2x+2hcos 2x+2h-sin 2xcos 2xh                   =limh0 sin 2x+2h cos 2x-cos 2x+2h sin 2x h cos 2x+2h cos 2x                   =limh0 sin 2x+2h-2xh cos 2x+2h cos 2x                   =1cos 2xlimh0 sin 2h 2h×2× limh01 cos 2x+2h                   =1cos 2x×2×1cos 2x                   =2cos22x                   =2 sec2 2x

iv ddxf(x)=limh0fx+h-fxh                    =limh0 tanx+h-tan xh×tanx+h+tan xtanx+h+tan x                   =limh0tanx+h-tan xhtanx+h+tan x                   =limh0sin x+hcos x+h-sin xcos xhtanx+h+tan x                   =limh0 sin x+h cos x-cos(x+h) sin x htanx+h+tan x cos x+h cos x                    =limh0 sin hhtanx+h+tan x cos x+h cos x                                =limh0 sin hhlimh0 1tanx+h+tan x cos x+h cos x                   =112 tan xcos2 x                   =sec2 x2 tan x

Page No 30.26:

Question 5:

(i) sin 2x
(ii) cos x
(iii) tan x
(iv) tan x2

Answer:

i Let f(x)= sin2x Thus, we have: f(x+h)=sin2x+hddxf(x)=limh0fx+h-fxh=limh0 sin 2x+2h-sin 2xhWe know:sin C- sin D=2 sin C-D2 cos C+D2=limh0 2 sin2x+2h-2x cos2x+2h-2x h=limh0 2×2 sin2x+2h-2x2 cos2x+2h+2x2 2h+2x-2x=limh0 2×2 sin2x+2h-2x2 cos2x+2h-2x2 2x+2h-2x2x+2h+2x=limh0 2×2 sin2x+2h-2x2  cos2x+2h-2x2 2× 2x+2h-2x22x+2h+2x= limh0 sin2x+2h-2x22x+2h-2x2limh0  2cos 2x+2h-2x22x+2h+2x   =1× 2cos2x22x                  limh0sin2x+2h-2x22x+2h-2x2=1=cos2x2x


(ii)  Let f(x) = cos x Thus, we have: f(x+h)=cos x+hddxfx=limh0f(x+h)-f(x) h= limh0cos x+h-cos xhWe know: cos C -cos D = -2sinC+D2 sinC-D2= limh0 -2sin x+h+x2 sinx+h-x2h= limh0 -2sin x+h+x2 sinx+h-x2x+h-x=limh0 -2sin x+h+x2 sinx+h-x22×x+h+xx+h-x2=limh0sinx+h-x2x+h-x2limh0 -sinx+h+x2x+h+x                       =1×-sinx2x                     limh0sinx+h-x2x+h-x2=1 =-sinx2x

(iii) Let f(x) = tanxThus, we have:(x+h)=tanx+hddx(f(x))=lim h0 f(x+h)-f(x) h=lim h0 tanx+h-tanx h=lim h0 sin x+h-xh cosx+h cos x           tan A-tan B= sin(A-B)cos A cos B      =lim h0      sin x+h-xx+h-x cosx+h cos x  =lim h0 sin x+h-xx+h-xx+h-xcosx+h cos x =lim h0sin x+h-xx+h-x.lim h0 1x+h+xcosx+hcosx           lim h0sinx+h-xx+h-x=1=1×12xcosxcosx=12xsec2x


(iv) Let f(x)= tan x2 Thus, we have:f(x+h) = tan (x+h)2ddxf(x)=limh0f(x+h)-f(x)h=limh0 tan (x+h)2-tanx2h=limh0sinx+h2-x2h cos x+h2cosx2              tan A-tan B=sin (A-B)cos A cos B=limh0 sin(x2+h2+2hx-x2)hcosx+h2cosx2=limh0sin(hh+2x)hh+2x cosx+h2cosx2×h+2x=limh0 sin(hh+2x)(hh+2x)limh0 h+2xcos(x+h)2cos x2         As limh0 sin(hh+2x)(hh+2x)=1=1×2xcos2 x2=2x sec2x2  



Page No 30.3:

Question 1:

Find the derivative of f (x) = 3x at x = 2

Answer:

We have:

f'(2) =lim                  h0f(2+h)-f(2)h        =limh03(2+h)-3(2)h        =limh06+3h-6h       =limh03hh       =3

Page No 30.3:

Question 2:

Find the derivative of f (x) = x2 − 2 at x = 10

Answer:

We have:
f'(x)=limh0f(10+h)-f(10)h       =limh0(10+h)2-2-(102-2)h      =limh0100+h2+20h-2-100+2h      =limh0h2+20hh      =limh0h(h+20)h     =limh0h+20    =0+20    =20

Page No 30.3:

Question 3:

Find the derivative of f (x) = 99x at x = 100

Answer:

We have:
f'(100)=limh0f(100+h)-f(100)h            =limh099(100+h)-99(100)h           =limh09900+99h-9900h          =limh099hh          =limh099           =99

Page No 30.3:

Question 4:

Find the derivative of f (x) x at x = 1

Answer:

We have:
f'(x)=limh0f(1+h)-f(1)h     =limh01+h-1h    =limh01    =1

Page No 30.3:

Question 5:

Find the derivative of f (x) = cos x at x = 0

Answer:

We have:
f'(x)=limh0f(0+h)-f(0)h=limh0f(h)-f(0)h=limh0cosh-cos0h=limh0cosh-1h=lim     h0 -2 sin2 h2h=limh0 -2 sin2 h2h24 ×h4==limh0 -1 ×h2=0

Page No 30.3:

Question 6:

Find the derivative of f (x) = tan x at x = 0

Answer:

We have:
f'(x)=limh0f(0+h)-f(0)h     =limh0f(h)-f(0)h    =limh0tanh-tan0h   =limh0tanhh   = 1

Page No 30.3:

Question 7:

Find the derivatives of the following functions at the indicated points:

(i) sin x at x = π2

(ii) x at x = 1

(iii) 2 cos x at x = π2

(iv) sin 2x at x = π2

Answer:

i We have:f'π2=limh0fπ2+h-fπ2h        =limh0sinπ2+h-sinπ2h      =limh0cos h-1h    =lim     h0 -2 sin2 h2h   =limh0 -2 sin2 h2h24 ×h4  =limh0-1 ×h2  =0
iiWe have:f'(x)=limh0f(1+h)-f(1)h      =limh01+h-1h    =limh01     =1

iii We have:f'π2=limh0fπ2+h-fπ2h          =limh02cosπ2+h-cosπ2h          =limh0-2sin h-0h          =-2limh0sinhh          =-2(1)          =-2
iv We have:f'π2=limh0fπ2+h-fπ2h          =limh0sin2π2+h-sin2π2h         =limh0sin(π+2h)-0h         =limh0-sin2hh×22             =limh0 -sin 2h2h×2         =-2



Page No 30.33:

Question 1:

x4 − 2 sin x + 3 cos x

Answer:

ddxx4-2 sin x+ 3 cos x=ddxx4-2ddx sin x+ 3ddxcos x=4x4-2 cos x - 3 sin x

Page No 30.33:

Question 2:

3x + x3 + 33

Answer:

ddx3x+x3+33=ddx3x+ddxx3+ddx33=3x log 3+3x2 + 0=3x log 3+3x2

Page No 30.33:

Question 3:

x33-2x+5x2

Answer:

ddxx33-2x+5x2=13ddxx3-2ddxx12+5ddxx-2=133x2-2.12.x-12+5-2x-3=x2-x-12-10x-3

Page No 30.33:

Question 4:

ex log a + ealong x + ea log a

Answer:

ddxex log a+ea log x+ea log a=ddxex log a+ddxea log x+ddxea log a=ddxelog ax+ddxelog xa+ddxelog aa=ddxax+ddxxa+ddxaa=ax log a + axa-1+0 =ax log a + axa-1

Page No 30.33:

Question 5:

(2x2 + 1) (3x + 2)

Answer:

ddx2x2+13x+2=ddx6x3+4x2+3x+2=6ddxx3+4ddxx2+3ddxx+ddx2=63x2+42x+31+0=18x2+8x+3

Page No 30.33:

Question 6:

 log3x + 3 logex + 2 tan x

Answer:

ddxlog3 x+ 3 loge x+2 tan x=ddxlog xlog 3  +3ddx loge x+2ddxtan x=1log 3.1x+3.1x+2 sec2 x=1x log 3+3x+2 sec2x



Page No 30.34:

Question 7:

x+1xx+1x

Answer:

ddxx+1xx+1x=ddxx+x-1x12+x-12=ddxx32+x12+x-12+x-32=ddxx32+ddxx12+ddxx-12+ddxx-32=32x12+12x-12-12x-32-32x-52

Page No 30.34:

Question 8:

x+1x3

Answer:

ddxx+1x3=ddxx3+3x21x+3x1x2+1x3=ddxx32+3ddxx12+3ddxx-12+ddxx-32=32x32-1+3.12x12-1+3-12x-12-1+-32x-32-1=32x12+32x-12-32x-32-32x-52 

Page No 30.34:

Question 9:

2x2+3x+4x

Answer:

ddx2x2+3x+4x=ddx2x2x+ddx3xx+ddx4x=2ddxx+3ddx1+4ddxx-1=21+30+4-1x-2=2-4x2

Page No 30.34:

Question 10:

(x3+1)(x-2)x2

Answer:

ddxx3+1x-2x2=ddxx4-2x3+x-2x2=ddxx4x2-2ddxx3x2+ddxxx2-ddx2x2=ddxx2-2ddxx+ddxx-1-2ddxx-2=2x-2-1x2-2-2x-3=2x-2-1x2+4x3

Page No 30.34:

Question 11:

a cos x+b sin x+csin x

Answer:

ddxa cos x+b sin x+csinx=ddxa cos xsin x+ddxb sin xsin x+ddxcsin x=addxcot x+ddxb+cddxcosec x=-a cosec2x+0-c cosec x cot x=-a cosec2x-c cosec x cot x

Page No 30.34:

Question 12:

2 sec x + 3 cot x − 4 tan x

Answer:

ddx2 sec x+ 3 cot x-4 tan x=2ddxsec x+3ddxcot x-4ddxtan x=2 sec x tan x-3 cosec2x-4 sec2x

Page No 30.34:

Question 13:

a0xn + a1xn−1 + a2xn2 + ... + an1x + an.

Answer:

ddxa0 xn+ddxa1 xn-1+ddxa2 xn-2+...+an-1 x+ an=a0ddxxn+a1ddxxn-1+a2ddxxn-2+...+an-1ddxx+ddxan=na0 xn-1+n-1 a1 xn-2+n-2a2 xn-3+....+an-11+0=na0 xn-1+n-1 a1 xn-2+n-2a2 xn-3+....+an-1

Page No 30.34:

Question 14:

1sin x+2x+3+4logx 3

Answer:

ddx1sin x+2x+3+4logx3=ddxcosec x+2x. 23+4log 3log x=ddxcosec x+23ddx2x+4log 3ddxlog x=-cosec x cot x+23.2x.log 2+4log 3.1x=-cosec x cot x+2x+3.log 2+4xlog 3

Page No 30.34:

Question 15:

(x+5)(2x2-1)x

Answer:

ddxx+52x2-1x=ddx2x3+10x2-x-5x=ddx2x3x+ddx10x2x-ddxxx-ddx5x=2ddxx2+10ddxx-ddx1-5ddxx-1=22x+101-0-5-1x-2=4x+10+5x2

Page No 30.34:

Question 16:

log1x+5xa-3ax+x23+6 x-34

Answer:

ddxlog 1x+5xa-3ax+x23+6 x-34=ddxlog x-12+5ddxxa-3ddxax+ddxx23+6ddxx-34=ddx-12log x+5ddxxa-3ddxax+ddxx23+6ddxx-34=-12.1x+5axa-1-3ax log a+23x-13+6-34x-74=-12x+5axa-1-3ax log a+23x-13-92x-74

Page No 30.34:

Question 17:

cos (x + a)

Answer:

ddxcos x+a=ddxcos x cos a - sin x sin a=cos addxcos x-sin a ddxsin x=-cos a sin x-sin a cos x=-sin x cos a + cos x sin a=-sinx+a

Page No 30.34:

Question 18:

cos (x-2)sin x

Answer:

ddxcosx-2sin x=ddxcos x cos 2+sin x sin 2sin x=ddxcos x cos 2sin x+ddxsin x sin 2sin x=cos 2ddxcot x+sin 2ddx1=cos 2 -cosec2x+sin 2 0=-cosec2x cos 2

Page No 30.34:

Question 19:

If y=sinx2+cosx22, find dydx at x=π6.

Answer:

dydx=ddxsin x2+cos x22       =ddxsin2 x2+cos2x2+2 sin x2cos x2      =ddx1+sinx      =ddx1+ddxsin x      =0+cos x      =cos xdydx at x = π6 = cos π6 = 32

Page No 30.34:

Question 20:

If y=2-3 cos xsin x, find dydx at x=π4

Answer:

dydx=ddx2-3 cos xsin x      =ddx2sin x-ddx3 cos xsin x      =2ddxcosec x-3ddxcot x      =-2 cosec x cot x+3 cosec2xdydx at x=π4=-2 cosec π4 cot π4+3 cosec2π4                      =-221+322                      =-22+6                      =6-22

Page No 30.34:

Question 21:

Find the slope of the tangent to the curve f (x) = 2x6 + x4 − 1 at x = 1.

Answer:

Slope of the tangent = f'(x)                                   =ddx2x6+x4-1                                   =2ddxx6+ddxx4-ddx1                                   =12x5+4x3 Slope of the tangent at x=1:1215+413=12+4=16

Page No 30.34:

Question 22:

If y=xa+ax, prove that 2xydydx=xa-ax

Answer:

y=xa+ax=1ax12+ax-12dydx=1a12x-12+a-12x-32LHS = 2xy dydx        = 2x 1ax12+ax-121a12x-12+a-12x-32       = 2x12a-12x+12x-a2x2        = 2x12a-a2x2        = xa-ax        = RHS Hence, proved.

Page No 30.34:

Question 23:

Find the rate at which the function f (x) = x4 − 2x3 + 3x2 + x + 5 changes with respect to x.

Answer:

Rate = f'(x)        =ddxx4-2x3+3x2+x+5        =ddxx4-2ddxx3+3ddxx2+ddxx+ddx5        =4x3-23x2+32x+1+0        =4x3-6x2+6x+1

Page No 30.34:

Question 24:

If y=2x93-57x7+6x3-x, find dydx at x=1.

Answer:

dydx=ddx2x93-57x7+6x3-x      =23ddxx9-57ddxx7+6ddxx3-ddxx      =239x8-577x6+63x2-1      =6x8-5x6+18x2-1dydx at x = 1: 618-516+1812-1=6-5+18-1=18

Page No 30.34:

Question 25:

If for f (x) = λ x2 + μ x + 12, f' (4) = 15 and f' (2) = 11, then find λ and μ.

Answer:

f'x=λddxx2+μddxx+ddx12f'x=2λx+μ       1Given: f'4=152λ4+μ=15     From 18λ+μ=15           2Also, given: f'2=112λ2+μ=11      From 14λ+μ=11                3Subtracting equation (3) from equation (2):4λ=4λ=1Substituting this in equation (3):41+μ=11μ=7∴ λ=1 and μ=7

Page No 30.34:

Question 26:

For the function f(x)=x100100+x9999+...+x22+x+1. Prove that f' (1) = 100 f' (0).

Answer:

f'x =ddxx100100+x9999+...+x22+x+1        =1100100 x99+19999 x98+...+122x+1+0        =x99+x98+...+x+1f'1=199+198+...+1+1        =99+1        =100f'0=0+0+...+0+1        =1RHS=100 f'0        =1001        =100        =f'1        =LHS f'1=100 f'0



Page No 30.39:

Question 1:

x3 sin x

Answer:

Let u=x3; v= sin xThen, u'= 3x2; v'=cos xUsing the product rule:ddxuv=uv'+vu'ddxx3 sin x=x3 cos x+ sin x 3x2                       =x2 x cos x+ 3 sin x

Page No 30.39:

Question 2:

x3ex

Answer:

Let u= x3; v=exThen, u'=3x2; v'=exUsing the product rule:ddxuv=uv'+vu'ddxx3 ex=x3 ex + ex 3x2                   =x2ex x+3

Page No 30.39:

Question 3:

x2ex log x

Answer:

Let u=x2; v=ex; w=log xThen, u'=2x; v'=ex, w= 1xUsing the product rule:ddxuvw=u'vw++uv'w+uvw'ddxx2ex log x=2x ex log x+x2 ex log x+x2 ex 1x                            =2x ex log x+x2 ex log x+x ex                            =x ex 2 log x+ x log x+1

Page No 30.39:

Question 4:

xn tan x

Answer:

Let u= xn; v=tan xThen, u'=nxn-1; v'=sec2xUsing the product rule:ddxuv=uv'+vu'ddxxn tan x=xn sec2x+tan xnxn-1                        =xn-1 x sec2x + n tan x

Page No 30.39:

Question 5:

xn logax

Answer:

Let u=xn; v=loga x=log xlog aThen, u'=n xn-1; v'=1x log aUsing the product rule:ddxuv=uv'+vu'ddxxn loga x=xn . 1x log a+loga x n xn-1                          =xn-11 log a+loga x n xn-1                          =xn-1 1log a+n loga x

Page No 30.39:

Question 6:

(x3 + x2 + 1) sin x

Answer:

Let u=x3+x2+1; v= sin xThen,u'= 3x2+2x; v'=cos xBy product rule,ddxuv=uv'+vu'ddxx3+x2+1 sin x=x3+x2+1 cos x+3x2+2x sin x 

Page No 30.39:

Question 7:

sin x cos x

Answer:

Let u= sin x; v= cos xThen, u'= cos x; v'= - sin xUsing the product rule:ddxuv=uv'+vu'ddxsin x cos x=sin x -sin x+cos x . cos x                             = -sin2x+cos2x                             = cos 2x

Page No 30.39:

Question 8:

2x cot xx

Answer:

2x cot xx=2x cot x x-12Let u=2x; v=cot x; w=x-12Then, u'=2x log 2; v'=-cosec2 x; w'=-12x-32Using the product rule:ddxuvw=u'vw+uv'w+uvw'ddx2x cot x x-12 =2x log 2.cot x.x-12+2x-cosec2 x x-12+2x cot x-12x-32                                     =2x log 2.cot x.1x+2x-cosec2 x1x+2x cot x-12xx                                      =2xxlog 2 . cot x-cosec2x-cot x2x

Page No 30.39:

Question 9:

x2 sin x log x

Answer:

Let u=x2; v=sin x; w=log xThen, u'=2x; v'=cos x; w'=1xUsing the product rule:ddxuvw=u'vw+uv'w+uvw'ddxx2 sin x log x=2x sin x log x+x2 cos x log x+x2 sin x.1x                                 =2x sin x log x+x2 cos x log x+x sin x

Page No 30.39:

Question 10:

x5ex + x6 log x

Answer:

ddxx5 ex + x6 log x=x5 ddxex+ex ddxx5+x6ddxlog x+ log x ddxx6=x5 ex +ex 5x4 +x6.1x+log x6x5=x5 ex +ex 5x4 +x5+log x6x5=x4 xex + 5ex +x+6x log x

Page No 30.39:

Question 11:

(x sin x + cos x) (x cos x − sin x)

Answer:

u= x sin x+cos x; v= x cos x-  sin xu'=x cos x+ sin x- sin x = x cos x ; v'=-  x sin x+cos x- cos x=-  x sin x             Using the product rule:ddxuv=uv'+vu'ddxx sin x+cos xx cos x-  sin x= x sin x+cos x-  x sin x + x cos x-  sin x x cos x                                                                   =-  x2 sin2 x-  x cos x sin x +x2 cos2 x-  x cos x sin x                                                                    =x2 cos2x -  sin2x-  x2 sin x cos x                                                                   =x2 cos 2x-  xsin 2x                                                                   = x x cos 2x- sin 2x

Page No 30.39:

Question 12:

(x sin x + cos x ) (ex + x2 log x)

Answer:

Let u=x sin x + cos x; v = ex + x2 log x Then, u' = xddxsin x+sinx  ddxx-sin x                  = x cos x +sinx - sinx                 = x cos x v' = ex +x2 ddxlog x + log x ddxx2       = ex +x + 2x log x                          Using the product rule: ddxuv= u v ' + v u'ddxx sinx + cos x ex + x2 cos x=x sin x + cos x ex +x + 2x log x   + ex + x2 log x  x cos x

Page No 30.39:

Question 13:

(1 − 2 tan x) (5 + 4 sin x)

Answer:

Let u=1-2 tan x; v = 5 + 4 sin x Then, u' = -2 sec2x; v' = 4 cos xUsing the product rule:ddxuv=u v' + v u'ddx1-2 tan x 5 + 4 sin x =1-2 tan x 4 cos x+ 5 + 4 sin x -2 sec2x= 4 cos x -8 ×sinxcosxcosx-10sec2x-8× sinxcos2x=4 cos x - 8 sin x - 10sec2 x-8 sec x tan x =4cos x- 2 sin x-52 sec2 x-2 sec x tan x

Page No 30.39:

Question 14:

(1 +x2) cos x

Answer:

Let u=1+x2; v= cos xThen ,u'=2x; v' =-sinxUsing the product rule:ddxuv=uv'+vu'ddx1+x2cos x=1+x2-sinx+ cos x2x                                   =-sinx - x2 sinx + 2x cos x 

Page No 30.39:

Question 15:

sin2x

Answer:

ddxsin2 x=2 sin x ddxsin x    (Using the chain rule)= 2 sin x cos x= sin 2x

Page No 30.39:

Question 16:

logx2x

Answer:

logx2 x=log xlog x2      (by change of base property)           =log x 2 log x        log x2=2 log x           =12Now ddxlogx2 x =ddx12                               =0           12 is a constant

Page No 30.39:

Question 17:

ex log x tan x

Answer:

Let u=ex; v= log x; w=tan xThen , u' = ex; v'= 1x×12x=12x; w' = sec2 xUsing the product rule:ddxuvw=u'vw + uv'w + uvw'                 =ex  log xtan x+ex×12xtan x+exlog x sec2 x                 =ex log x12.tan x+tanx2x+log x12.sec2 x                 =ex 12 log x. tan x+tanx2x+ 12 log x. sec2 x                  =ex2log x.tan x+tan xx+log x.sec2x

Page No 30.39:

Question 18:

x3ex cos x

Answer:

Let u=x3; v=ex; w=cos xThen ,u'=3x2; v'=ex; w'=-sin xUsing the product rule:ddxuvw=u'vw+uv'w+uvw'ddxx3 ex cos x=3x2 ex cos x+x3 ex cos x+x3 ex -sin x                              =x2 ex 3 cos x+x cos x-x sin x

Page No 30.39:

Question 19:

x2 cosπ4sin x

Answer:

x2 cos π4sin x=x2 cos π4 cosec xLet u=x2; v=cos π4; w= cosec xThen, u'=2x; v'=0; w'=-cosec x cot xUsing the product rule:ddxuvw=u'vw+uv'w+uvw'ddxx2 cos π4 cosec x=2x cos π4cosec x+ x2 .0.cosec x+x2 cos π4-cosec x cot x                                           = cos π42x cosec x -x2 cosec x cot x                                           = cos π42xsin x -x2  cot xsin x

Page No 30.39:

Question 20:

x4 (5 sin x − 3 cos x)

Answer:

Let u=x4; v=5 sin x - 3 cos xThen, u'=4x3; v'= 5 cos x - 3 (-sinx) = 5 cos x + 3 sin x According to the product rule:ddxuv=u v'+v u'ddxx45 sin x - 3 cos x=x4  5 cos x + 3 sin x +5 sin x - 3 cos x 4x3                                              =x35x cos x + 3 x sin x + 20 sin x - 12 cos x                                              =x33x+20 sin x +5x - 12 cos x                                              =3 x4 sinx +20x3sin x+5x cos x -12 cos x 

Page No 30.39:

Question 21:

(2x2 − 3) sin x

Answer:

Let u=2x2-3; v=sin xThen,u'=4x; v'= cos xUsing the product rule:ddxuv=uv'+vu'ddx2x2-3 sin x =2x2-3 cos x+ 4x sin x 

Page No 30.39:

Question 22:

x5 (3 − 6x−9)

Answer:

Let u=x5; v=3-6x-9Then, u'=5x4; v'=54x-10Using the product rule:ddxuv=uv'+vu'ddxx53-6x-9=x554x-10+3-6x-95x4                               =54 x-5+15 x4-30x-5                               =15 x4+24x-5

Page No 30.39:

Question 23:

x4 (3 − 4x−5)

Answer:

Let u=x-4; v=3-4x-5Then, u'=-4x-5; v'=20 x-6Using the product rule:ddxuv=uv'+vu'ddxx-4 3-4x-5=x-4 20 x-6+3-4x-5-4x-5                                  =20x-10-12x-5+16x-10                                  =-12x-5 +36x-10

Page No 30.39:

Question 24:

x−3 (5 + 3x)

Answer:

Let u=x-3; v=5+3xThen, u=-3x-4; v'=3Using the product rule:ddxuv=uv'+vu'ddxx3 5+3x=x-3.3+5+3x -3x-4                            =3x-3 - 15 x-4 - 9 x-3                            =-15 x-4-6 x-3

Page No 30.39:

Question 25:

Differentiate in two ways, using product rule and otherwise, the function (1 + 2 tan x) (5 + 4 cos x). Verify that the answers are the same.

Answer:

Product rule (1st method):Let u=1+2 tan x; v=5+4 cos xThen, u'=2sec2x; v'=-4 sin xUsing the product rule:ddxuv=uv'+vu'ddx1+2 tan x5+4 cos x=1+2 tan x-4 sin x+5+4 cos x2sec2x                                                   =-4 sin x-8 tan x sin x+10 sec2x+8 sec x                                                   =-4 sin x+10 sec2x+8cosx- 8sin2xcos x                                                   = -4 sin x+10 sec2x+81-sin2xcos x                                                   =-4 sin x+10 sec2x+8cos2xcos x                                                   =-4 sin x+10 sec2x+8 cos x2nd method:1+2 tan x5+4 cos x=5+4 cos x+10 tan x+ 8 sin xNow, we have:ddx1+2 tan x5+4 cos x=ddx5+4 cos x+10 tan x+ 8 sin x                                                    =-4 sin x+10 sec2x+8 cos xUsing both the methods, we get the same answer.

Page No 30.39:

Question 26:

Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.
(i) (3x2 + 2)2
(ii) (x + 2) (x + 3)
(iii) (3 sec x − 4 cosec x) (−2 sin x + 5 cos x)

Answer:

i Product rule (1st method):Let u=3x2+2; v=3x2+2Then, u'=6x; v'=6xUsing the product rule:ddxuv=uv'+vu'ddx3x2+23x2+2=3x2+26x+3x2+26x                                      =18x3+12x+18x3+12x                                      =36x3+24x  2nd method:ddx3x2+22=ddx9x4+12x2+4                          =36x3+24xUsing both the methods, we get the same answer.

ii Product rule (1st method):Let u=x+2; v=x+3Then, u'=1; v'=1Using the product rule:ddxuv=uv'+vu'ddxx+2x+3=x+21+x+31                               =x+2+x+3                               =2x+52nd method:ddxx+2x+3=ddxx2+5x+6                                =2x+5Using both the methods, we get the same answer.

iii Product rule (1st method):Let u=3 sec x-4 cosec x; v=-2 sin x+ 5 cos xThen, u'=3 sec x tan x+4 cosec x cot x; v'=- 2 cos x-5 sin xUsing the product rule:ddxuv=uv'+vu'ddx3 sec x-4 cosec x-2 sin x+ 5 cos x=3 sec x-4 cosec x- 2 cos x-5 sin x+-2 sin x+ 5 cos x3 sec x tan x+4 cosec x cot x                                                                                 =-6+15 tan x+8  cot x+20 -6 tan2x-8 cot x-15 tan x+ 20  cot2x                                                                                 =-6+20 -6sec2x -1+ 20  cosec2x-1                                                                                 =-6+20 -6sec2x +6+ 20 cosec2x-20                                                                                 =-6 sec2x+20 cosec2x2nd method:ddx3 sec x-4 cosec x-2 sin x+ 5 cos x=ddx-6 sec x sin x +15 sec x cos x+8 cosec x sin x -20 cosec x cos x                                                                                 =ddx-6 sin xcos x +15cos xcos x+8 sin xsin x -20 cos xsin x                                                                                 =ddx-6 tan x+15 +8 -20 cot x                                                                                 =ddx-6tan x-20 cot x+23                                                                                 =-6 sec2x+20 cosec2xUsing both the methods, we get the same answer.

Page No 30.39:

Question 27:

(ax + b) (a + d)2

Answer:

(ax+b)(a+d)2Let u =ax+b, v= a+d2Then, u'=a, v'= 0Using the product rule:ddxuv=u v' + v u'ddx(ax+b)(a+d)2=(ax+b)×0+ a+d2×a ddx(ax+b)(a+d)2=aa+d2

Page No 30.39:

Question 28:

(ax + b)n (cx + d)n

Answer:

ax+bn cx+dnLet u =ax+bn, v=cx+dnThen, u'=naax+bn-1, v'=nccx+dn-1Using the product rule:ddxuv=uv'+u'vddxax+bn cx+dn=ax+bn×nccx+dn-1+naax+bn-1×cx+dn                                          =nax+bn-1cx+dn-1acx+cb+acx+ad                                          =nax+bn-1cx+dn-12acx+cb+ad



Page No 30.44:

Question 1:

x2+1x+1

Answer:

Let u=x2+1; v=x+1Then, u'=2x; v'=1Using the quotient rule:ddxuv=vu'-uv'v2ddxx2+1x+1=x+12x-x2+11(x+1)2                       =2x2+2x-x2-1(x+1)2                       =x2+2x-1(x+1)2

Page No 30.44:

Question 2:

2x-1x2+1

Answer:

Let u=2x-1; v=x2+1;Then, u'=2; v'=2xUsing the quotient rule:ddxuv=vu'-uv'v2ddx2x-1x2+1=x2+12-2x-12x(x2+1)2                        =2x2+2-4x2+2x(x2+1)2                        =-2x2+2x+2(x2+1)2                        =21+x-x2(x2+1)2

Page No 30.44:

Question 3:

x+ex1+log x

Answer:

Let u=x+ex; v=1+logxThen, u'=1+ex; v'=1xUsing the quotient rule:ddxuv=vu'-uv'v2ddxx+ex1+logx=1+log x1+ex-x+ex1x(1+log x)2                           =x+xex+x log x+x log x ex-x-exx(1+logx)2                           =x log x+ x log x ex-ex+x exx(1+logx)2                           =x log x 1+ex-ex1-xx(1+logx)2

Page No 30.44:

Question 4:

ex-tan xcot x-xn

Answer:

Let u=ex-tan x; v=cot x-xnThen, u'=ex-sec2x; v'=-cosec2x-nxn-1Using the quotient rule:ddxuv=vu'-uv'v2ddxex-tan xcot x-xn=cot x-xnex-sec2x-ex-tan x-cosec2x-nxn-1cot x-xn2                              =cot x-xnex-sec2x+ex-tan xcosec2x+nxn-1cot x-xn2

Page No 30.44:

Question 5:

ax2+bx+cpx2+qx+r

Answer:

Let u=ax2+bx+c; v=px2+qx+rThen, u'=2ax+b; v'=2px+qUsing the quotient rule:ddxuv=vu'-uv'v2ddxax2+bx+cpx2+qx+r=px2+qx+r2ax+b-ax2+bx+c2px+qpx2+qx+r2                                  =2apx3+2aqx2+2arx+bpx2+bqx+br-2apx3-2bpx2-2pcx-aqx2-bqx-cqpx2+qx+r2                                  =aq-bpx2+2ar-xpx+br-cqpx2+qx+r2

Page No 30.44:

Question 6:

x1+tan x

Answer:

Let u=x; v=1+tan xThen, u'=1; v'= sec2xUsing the quotient rule:ddxuv=vu'-uv'v2                =1+tan x×1-xsec2x1+tanx2                =1+tan x-xsec2x1+tanx2

Page No 30.44:

Question 7:

1ax2+bx+c

Answer:

ddx1ax2+bx+c=ddxax2+bx+c-1=-1ax2+bx+c-2ddxax2+bx+c    (Using the chain rule)=-1ax2+bx+c-22ax+b=-2ax+bax2+bx+c2

Page No 30.44:

Question 8:

ex1+x2

Answer:

Let u=ex; v=1+x2Then, u'=ex; v'=2xUsing the chain rule:ddxuv=vu'-uv'v2ddxex1+x2=1+x2ex-ex2x1+x22                       =ex+x2ex-2xex1+x22                       =ex1+x2-2x1+x22                       =ex1-x21+x22

Page No 30.44:

Question 9:

ex+sin x1+log x

Answer:

Let u=ex+sin x; v=1+ log xThen, u'=ex+cos x; v'= 1xUsing the quotient rule:ddxuv=vu'-uv'v2ddxex+sin x1+ log x=1+log xex+cos x-ex+sin x 1x1+ log x2                              =x1+log xex+cos x-ex+sin xx1+ log x2

Page No 30.44:

Question 10:

x tan xsec x+tan x

Answer:

Let u=x tan x; v=sec x+tan xThen, u'=x sec2x+tan x; v'=sec x tan x + sec2xUsing the quotient rule:ddxuv=vu'-uv'v2ddxxtan x sec x +tan x =sec x+tan xx sec2x+tan x-x tan xsec x tan x + sec2xsec x+tan x2                                     =x sec3x+x sec2xtanx+secx tanx+tan2x-x sec x tan2x-x tanx sec2xsec x+tan x2                                     =sec x+tan xx sec2x+tan x-x tan x secxsec x+tan xsec x+tan x2                                     =x sec2x+tan x-x tanx secxsec x+ tan x                                       =x sec xsec x -tan x+tan xsec x + tan x

Page No 30.44:

Question 11:

x sin x1+cos x

Answer:

Let u=x sinx; v=1+ cos xThen, u' = x cos x + sinx; v' = -sin xUsing the quotient rule:ddxuv=vu'-uv'v2ddxx sin x1+cos x=1+ cos xx cos x + sinx-x sinx-sin x1+ cos x2                             =1+ cos xx cos x + sinx+x sin2 x1+ cos x2                             =1+ cos xx cos x + sinx+x 1-cos2x1+ cos x2                             =1+ cos xx cos x + sinx+x1+cos x1-cos x1+ cos x2                             =1+ cos xx cos x + sinx+x- xcosx1+ cos x2                             =1+ cos xx+sin x1+ cos x2

Page No 30.44:

Question 12:

2x cot xx

Answer:

Let u=2xcot x; v=xThen, u'=-2xcosec2x+2xlog 2 cot x; v'=12xddxuv=vu'-uv'v2ddx2xcot xx=x-2xcosec2x+2xlog 2 cot x-2xcot x12xx2                          =x-2xcosec2x+2xlog 2 cot x-2x-1cot xxx                          =2x-xcosec2x+x cot x log 2 -12cot xxx                          =2x-xcosec2x+x cot x log 2 -12cot xx32

Page No 30.44:

Question 13:

sin x-x cos xx sin x+cos x

Answer:

Let u=sin x-x cos x; v=x sinx + cos xThen, u'=cosx+x sin x-cosx; v'=x cos x+sin x-sinx   =x sin x                           =x cos xUsing the quotient rule:ddxuv=vu'-uv'v2ddxsin x-x cos xx sinx + cos x=x sinx + cos xx sinx -sin x-x cos xx cos xx sinx + cos x2                                       =x2sin2x+x cosx sinx -x cos x sin x+x2cos2xx sinx + cos x2                                       =x2sin2x+cos2xx sinx + cos x2                                       =x2x sinx + cos x2

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Question 14:

x2-x+1x2+x+1

Answer:

Let u=x2-x+1; v= x2+x+1Then, u'=2x-1; v'=2x+1By quotient rule,ddxuv=vu'-uv'v2ddxx2-x+1x2+x+1=x2+x+12x-1-x2-x+12x+1x2+x+12                              =2x3+2x2+2x-x2-x-1-2x3+2x2-2x-x2+x-1x2+x+12                             =2x2-2x2+x+12                             =2x2-1x2+x+12

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Question 15:

a+xa-x

Answer:

Let u=a+x; v=a-xThen,u'=12x; v'=-12xUsing the quotient rule:ddxuv=vu'-uv'v2ddxa+xa-x=a-x12x-a+x-12xa-x2                              =a-x+a+x2xa-x2                              =2a2xa-x2                             =axa-x2

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Question 16:

a+sin x1+a sin x

Answer:

Let us use the quotient rule here.
We have:
u = a + sin x and v =1 + a sin x
u' = cos x and v'=a cos x

Using the quotient rule:ddxuv=vu'-uv'v2ddxa+sinx1+asinx=(1+asinx)(cosx)-(a+sinx)(acosx)(1+asin x)2                             =cosx+asinx cosx -a2cosx-a sinx cosx(1+asin x)2                             =cosx-a2cosx(1+asinx)2                            =(1-a2)cosx(1+a sinx)2

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Question 17:

10xsin x

Answer:

Let u=10x; v=sin xThen,u'=10xlog 10; v'= cos xUsing the quotient rule:ddxuv=vu'-uv'v2ddx10xsin x=sin x 10xlog 10-10x cos xsin2x                      =sin x 10xlog 10sin2x-10x cos xsin2x                      =10xlog 10 cosec x -10xcosec x cot x                      =10x cosec xlog 10 - cot x

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Question 18:

1+3x1-3x

Answer:

Let u=1+3x; v=1-3xThen,u'=3x log 3; v'=-3x log 3Using the quotient rule:ddxuv=vu'-uv'v2ddx1+3x1-3x=1-3x3x log 3-1+3x-3x log 31-3x2                       =3x log 3-32x log 3+3x log 3+32x log 31-3x2                       =2 . 3x log 31-3x2

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Question 19:

3xx+tan x

Answer:

Let u=3x; v=x+tan xThen, u'=3xlog 3; v'=1+sec2xBy quotient rule, we have:ddxuv=vu'-uv'v2ddx3xx+tan x=x+tan x3xlog 3-3x 1+sec2xx+tan x2                            =3xx+tan x log 3- 1+sec2xx+tan x2

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Question 20:

1+log x1-log x

Answer:

Let u=1+log x; v=1- log xThen, u'=1x; v'=-1xUsing the quotient rule:ddxuv=vu'-uv'v2ddx1+log x1- log x=1- log x1x-1+log x-1x1- log x2                             =1-log x+1+log xx1- log x2                             =2 x1- log x2

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Question 21:

4x+5 sin x3x+7 cos x

Answer:

Let u=4x + 5 sin x; v=3x + 7 cos xThen, u'=4+5 cos x; v'=3-7 sin xUsing the quotient rule:ddxuv=vu'-uv'v2ddx4x + 5 sin x3x + 7 cos x=3x + 7 cos x4+5 cos x-4x + 5 sin x3-7 sin x3x + 7 cos x2                                    =12x+15 x cos x+28 cos x+35 cos2x-12x +28 x sin x-15 sin x +35 sin2x3x + 7 cos x2                                    =15 x cos x+ 28 x sin x+28 cos x15 sinx +35sin2x+cos2x3x + 7 cos x2                                    =15 x cos x+ 28 x sin x+28 cos x15 sinx +353x + 7 cos x2

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Question 22:

x1+tan x

Answer:

Let u=x; v=1+tan xThen, u'=1; v'= sec2xUsing the quotient rule:ddxuv=vu'-uv'v2ddxx1+tan x=1+tan x1-xsec2x1+tan x2                            =1+tan x-xsec2x1+tan x2

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Question 23:

a+b sin xc+d cos x

Answer:

Let u=a+b sin x; v=c+d cos xThen, u'=b cos x; v'=-d sin xUsing the quotient rule:ddxuv=vu'-uv'v2ddxa+b sin xc+d cos x=c+d cos xb cos x-a+b sin x-d sin xc+d cos x2                                 =bc cos x+bd cos2x+ad sin x+bd sin2xc+d cos x2                                  =bc cos x+ad sin x+bd sin2x+cos2xc+d cos x2                                  =bc cos x+ad sin x+bd c+d cos x2

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Question 24:

px2+qx+rax+b

Answer:

Let u=px2+qx+r; v=ax+bThen, u'=2px+q; v'=aUsing the quotient rule:ddxuv=vu'-uv'v2ddxpx2+qx+rax+b=ax+b2px+q-px2+qx+raax+b2                                 =2ap x2+aq x+2bp x+bq-ap x2-aq x-arax+b2                                 =ap x2+2bp x+bq -arax+b2

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Question 25:

sec x-1sec x+1

Answer:

Let u=sec x-1; v=sec x+1Then,u'=sec x tan x; v'=sec x tan xUsing the quotient rule:ddxuv=vu'-uv'v2ddxsec x-1sec x+1=sec x+1sec x tan x-sec x-1sec x tan xsec x+12                             =sec2x tan x+ sec x tan x-sec2x tan x+sec x tan xsec x+12                             =2sec x tan xsec x+12

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Question 26:

x5-cos xsin x

Answer:

Let u=x5-cos x; v=sin xThen, u'= 5x4+sin x; v'=cos xUsing the quotient rule:ddxuv=vu'-uv'v2ddxx5-cos xsin x=sin x 5x4+sin x-x5-cos xcos xsin2x                               =-x5cos x + 5x4sin x+sin2x+cos2xsin2x                               =-x5cos x + 5x4sin x+1sin2x

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Question 27:

x+cos xtan x

Answer:

Let u=x+cos x; v=tan xThen, u'=1-sin x; v'= sec2xUsing the quotient rule:ddxuv=vu'-uv'v2ddxx+cos xtan x=tan x1-sin x-x+cos xsec2xtan2x

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Question 28:

xsinn x

Answer:

Let u=x; v=sinnxThen,u'=1; v'=n sinn-1x.cos xUsing the quotient rule:ddxuv=vu'-uv'v2ddxxsinnx=sinnx.1-x n sinn-1 x.cos xsinnx2                      =sinn-1xsinx-nx.cos xsin2n x                      =sinx-nx.cos xsin2n-n-1 x                      =sinx-nxcos xsinn+1 x

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Question 29:

ax+bpx2+qx+r

Answer:

Let u=ax+b; v=px2+qx+rThen,u'=a; v'=2px+qUsing the quotient rule:ddxuv=vu'-uv'v2ddxax+bpx2+qx+r=px2+qx+ra-ax+b2px+qpx2+qx+r2                                 =ap x2+aq x+ar-2ap x2-2bp x-aq x-bqpx2+qx+r2                                 =-apx2-2bp x+ar-bqpx2+qx+r2

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Question 30:

1ax2+bx+c

Answer:

Let u=1; v=ax2+bx+cThen,u'=0; v'=2ax+bUsing the quotient rule:ddxuv=vu'-uv'v2ddx1ax2+bx+c=ax2+bx+c0-12ax+bax2+bx+c2                                  =-2ax+bax2+bx+c2



Page No 30.46:

Question 1:

Write the value of limxcf(x)-f(c)x-c.

Answer:

Using the definition of derivative, we have:limxcfx-fxx-c=f'c

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Question 2:

Write the value of limxax f (a)-a f (x)x-a.

Answer:

limxaxfa-afxx-a=limxaxfa-afx-xfx+xfxx-a=limxaxfa-xfx+xfx-afxx-a=limxa-xfx-fa+x-afxx-a=limxa-xlimxafx-fax-a+limxax-afxx-a=-a f'a+f(a)



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Question 3:

If x < 2, then write the value of ddx(x2-4x+4).

Answer:

Given: x<2∴ 2-x>0ddxx2-4x+4=ddx2-x2=ddx2-x   (∵ 2-x>0)=0-1=-1

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Question 4:

If π2 < x < π, then find ddx1+cos 2x2.

Answer:

1+cos 2x2=2 cos2x2=cos2x=-cos x    (∵ π2<x<π)ddx1+cos 2x2=ddx-cosx=--sinx=sinx

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Question 5:

Write the value of ddx x x .

Answer:

Case 1:x>0|x|=xThus, we have: ddxx|x|=ddxx.x=ddxx2=2x           1Case 2:x<0|x|=-xThus, we have: ddxx|x|=ddxx.-x=ddx-x2=-2x           2From (1) and (2), we have:ddxx|x|=2x, if x>0-2x, if x<0

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Question 6:

Write the value of ddx x+x x.

Answer:

Case 1:x>0x=xx+xx=x+xx=2x2ddxx+xx=ddx2x2=4x         1Case 2:x<0x=-xx+xx=x-xx=0ddxx+xx=ddx0=0            2From (1) and (2), we have:ddxx+xx=4x, if x>00, if x<0

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Question 7:

If f (x) = |x| + |x−1|, write the value of ddxf (x).

Answer:

fx=x+x-1Case 1: x<0 (∴ x-1<-1<0)x=-x; x-1=-x-1=-x+1fx=-x+-x+1=-2xf'x=-2Case 2: 0< x <1 (∴ x>0 and x-1<0)x=x; x-1=-x-1=1-xfx=x+1-x=1f'x=0Case 3: x>1 ∴ x>1>0 ⇒ x>0)x=x; x-1=x-1fx=x+x-1=2x-1f'x=2From case 1, case 2 and case 3, we have:f'x=-2, when x<0    0, when 0<x<12, when x>1

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Question 8:

Write the value of the derivative of f (x) = |x − 1| + |x − 3| at x = 2.

Answer:

Let x = 2We know: 2>1 and 2<3∴ x>1 and x<3x-1=x-1 and x-3=-x-3=-x+3fx=x-1+x-3=x-1-x+3=2f'x=0

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Question 9:

If f (x) = x2x, write ddxf (x).

Answer:

Case 1: x>0x=xfx=x2x=x2x=xf'x=1Case 2: x<0x=-xfx=x2x=x2-x=-xf'x=-1From case 1 and case 2, we have:f'x=1, if x>0-1, if x<0

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Question 10:

Write the value of ddxlog x.

Answer:

Case 1: x>0x=x   ...1ddxlog x=log x                  =1x                  =1x    (from (1))Case 2: x<0x=-x     ...2ddxlog x=log -x                   =1-x                   =1x  (from (2))From case (1) and case(2), ddxlog x=1x

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Question 11:

If f (1) = 1, f' (1) = 2, then write the value of limx1f (x)-1x-1.

Answer:

limx1 fx-1x-1=limx1 fx-1x-1×fx+1fx+1×x+1x+1=limx1 fx-1x+1x-1fx+1=limx1 fx-1x-1× limx1x+1fx+1=f'1×1+1f1+1=2×21+1=2

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Question 12:

Write the derivative of f (x) = 3 |2 + x| at x = −3.

Answer:

Let x = -3We know:-3<-2Thus, we have: x<-2It gives x+2<0. 2+x=x+2=-x+2=-x-2fx=3 2+x=3-x-2=-3x-6f'x=-3ddxx-ddx6=-3

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Question 13:

If |x| < 1 and y = 1 + x + x2 + x3 + ..., then write the value of dydx.

Answer:

The given series is a geometric series where a = 1 and r = x.
fx=1+x+x2+x3+...=11-xSum of the infinite series of a geometric series is a1-r.f'x=-1(1-x)2ddx(1-x)=-1(1-x)2(-1)=1(1-x)2

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Question 14:

If f (x) = logx2x3, write the value of f' (x).

Answer:

f(x)=logx2x3      =logx3logx2   (Change of base property)      =3 logx2 log x      =32f'x=0 (Since 32 is a constant)

Page No 30.47:

Question 1:

Mark the correct alternative in each of the following:

Let f(x) = x − [x], xR, then f'12 is

(a) 32                                  (b) 1                                  (c) 0                                  (d) −1

Answer:

Given: f(x) = x − [x], xR

Now,

For 0 ≤ x < 1, [x] = 0.

f(x) = x − 0 = x, ∀ x ∈ [0, 1)

Differentiating both sides with respect to x, we get

f '(x) = 1, ∀ x ∈ [0, 1)

 f'12=1

Hence, the correct answer is option (b).

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Question 2:

Mark the correct alternative in each of the following:

If fx=x-42x, then f '(1) is

(a) 54                                  (b) 45                                  (c) 1                                  (d) 0

Answer:


fx=x-42x      =12x-2x      =12x12-2x-12

Differentiating both sides with respect to x, we get

f'x=12×12x12-1-2×-12x-12-1                    fx=xnf'x=nxn-1f'x=14x-12+x-32f'1=14×1+1=54

Hence, the correct answer is option (a).

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Question 3:

Mark the correct alternative in each of the following:

If y=1+x1!+x22!+x33!+..., then dydx=

(a) y + 1                                 (b) y − 1                                  (c) y                                 (d) y2

Answer:

y=1+x1!+x22!+x33!+...

Differentiating both sides with respect to x, we get

dydx=ddx1+x1!+x22!+x33!+...       =ddx1+ddxx1!+ddxx22!+ddxx33!+ddxx44!+...       =ddx1+11!ddxx+12!ddxx2+13!ddxx3+14!ddxx4+...       =0+11!×1+12!×2x+13!×3x2+14!×4x3+...                                    y=xndydx=nxn-1
       =1+x1!+x22!+x33!+...                                                          nn!=1n-1!=y

dydx=y

Hence, the correct answer is option (c).



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Question 4:

Mark the correct alternative in each of the following:

If fx=1-x+x2-x3+...-x99+x100, then f'1 equals

(a) 150                                  (b) −50                                  (c) −150                                  (d) 50

Answer:

fx=1-x+x2-x3+...-x99+x100

Differentiating both sides with respect to x, we get

f'x=ddx1-x+x2-x3+...-x99+x100       =ddx1-ddxx+ddxx2-ddxx3+...-ddxx99+ddxx100       =0-1+2x-3x2+...-99x98+100x99       =-1+2x-3x2+...-99x98+100x99

Putting x = 1, we get

f'1=-1+2-3+...-99+100        =-1+2+-3+4+-5+6+...+-99+100        =1+1+1+ ... +1   50 terms        =50

Hence, the correct answer is option (d).

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Question 5:

Mark the correct alternative in each of the following:

If y=1+1x21-1x2, then dydx=

(a) -4xx2-12                                  (b) -4xx2-1                                  (c) 1-x24x                                  (d) 4xx2-1

Answer:


y=1+1x21-1x2  =x2+1x2-1

Differentiating both sides with respect to x, we get

dydx=x2-1×ddxx2+1-x2+1×ddxx2-1x2-12                Quotient rule       =x2-1×2x+0-x2+1×2x-0x2-12       =2x3-2x-2x3-2xx2-12       =-4xx2-12

Hence, the correct answer is option (a).

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Question 6:

Mark the correct alternative in each of the following:

If y=x+1x, then dydx at x = 1 is

(a) 1                                  (b) 12                                  (c) 12                                  (d) 0

Answer:


y=x+1x  =x12+x-12

Differentiating both sides with respect to x, we get

dydx=ddxx12+x-12       =ddxx12+ddxx-12       =12x12-1+-12x-12-1                               y=xndydx=nxn-1       =12x-12-12x-32

Putting x = 1, we get

dydxx=1=12×1-12×1=0

Thus, dydx at x = 1 is 0.

Hence, the correct answer is option (d).

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Question 7:

Mark the correct alternative in each of the following:

If fx=x100+x99+ ... +x+1, then f'1 is equal to

(a) 5050                                  (b) 5049                                  (c) 5051                                  (d) 50051

Answer:


fx=x100+x99+ ... +x+1

Differentiating both sides with respect to x, we get

f'x=ddxx100+x99+ ... +x+1       =ddxx100+ddxx99+ ... +ddxx2+ddxx+ddx1       =100x99+99x98+ ... +2x+1+0                                                       y=xndydx=nxn-1       =100x99+99x98+ ... +2x+1

Putting x = 1, we get

f'1=100+99+98+ ... +2+1        =100100+12                          Sn=nn+12        =50×101        =5050

Hence, the correct answer is option (a).

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Question 8:

Mark the correct alternative in each of the following:

If fx=1+x+x22+ ... +x100100, then f'1 is equal to

(a) 1100                                  (b) 100                                  (c) 50                                  (d) 0

Answer:


fx=1+x+x22+ ... +x100100

Differentiating both sides with respect to x, we get

f'x=ddx1+x+x22+ ... +x100100        =ddx1+ddxx+ddxx22+ ... +ddxx100100        =ddx1+ddxx+12ddxx2+ ... +1100ddxx100        =0+1+12×2x+ ... +1100×100x99                                 y=xndydx=nxn-1            =1+x+x2+ ... +x99

Putting x = 1, we get

f'1=1+1+1+ ... +1   100 terms        =100

Hence, the correct answer is option (b).

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Question 9:

Mark the correct alternative in each of the following:

If y=sinx+cosxsinx-cosx, then dydx at x = 0 is

(a) −2                                 (b) 0                                  (c) 12                                  (d) does not exist

Answer:

y=sinx+cosxsinx-cosx

Differentiating both sides with respect to x, we get

dydx=sinx-cosx×ddxsinx+cosx-sinx+cosx×ddxsinx-cosxsinx-cosx2                    Quotient rule      =sinx-cosx×ddxsinx+ddxcosx-sinx+cosx×ddxsinx-ddxcosxsinx-cosx2      =sinx-cosxcosx-sinx-sinx+cosxcosx+sinxsinx-cosx2      =-cos2x+sin2x-2cosx sinx-sin2x+cos2x+2sinx cosxsinx-cosx2
      =-1+2cosx sinx-1-2sinx cosxsinx-cosx2=-2sinx-cosx2

Putting x = 0, we get

dydxx=0=-2sin0-cos02=-20-12=-2

Thus, dydx at x = 0 is −2.

Hence, the correct answer is option (a).

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Question 10:

Mark the correct alternative in each of the following:

If y=sinx+9cosx, then dydx at x = 0 is

(a) cos 9                                  (b) sin 9                                  (c) 0                                  (d) 1

Answer:


y=sinx+9cosx

Differentiating both sides with respect to x, we get

dydx=cosx×ddxsinx+9-sinx+9×ddxcosxcos2x                Quotient rule       =cosx×cosx+9-sinx+9×-sinxcos2x       =cosx+9cosx+sinx+9sinxcos2x       =cosx+9-xcos2x       =cos9cos2x

Putting x = 0, we get

dydxx=0=cos9cos20=cos9                        (cos 0 = 1)

Thus, dydx at x = 0 is cos 9.

Hence, the correct answer is option (a).

Page No 30.48:

Question 11:

Mark the correct alternative in each of the following:

If fx=xn-anx-a, then f'a is

(a) 1                                  (b) 0                                  (c) 12                                  (d) does not exist

Answer:


Given: fx=xn-anx-a

Now, f(x) is not defined at x = a. Therefore, f(x) is not differentiable at x = a.

So, f'a does not exist.

Hence, the correct answer is option (d).

Page No 30.48:

Question 12:

Mark the correct alternative in each of the following:

If f(x) = x sinx, then f'π2= 

(a) 0                                  (b) 1                                  (c) −1                                  (d) 12

Answer:


f(x) = x sinx

Differentiating both sides with respect to x, we get

f'x=x×ddxsinx+sinx×ddxx                            Product rule       =x×cosx+sinx×1       =x cosx+sinx

Putting x=π2, we get

f'π2=π2×cosπ2+sinπ2            =π2×0+1            =1

Hence, the correct answer is option (b).



View NCERT Solutions for all chapters of Class 14