RD Sharma XII Vol 1 2017 Solutions for Class 12 Science Math Chapter 14 Differentials, Errors And Approximations are provided here with simple step-by-step explanations. These solutions for Differentials, Errors And Approximations are extremely popular among class 12 Science students for Math Differentials, Errors And Approximations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 1 2017 Book of class 12 Science Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 1 2017 Solutions. All RD Sharma XII Vol 1 2017 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 14.10:

Question 10:

Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2.

Answer:


Let: x=2x+x=2.01x=0.01fx=4x2+5x+2fx=2=16+10+2=28Now, y=fxdydx=8x+5 dy=y=dydxdx=8x+5×0.01=16+5×0.01=0.21 f2.01=y+y=28.21

Page No 14.10:

Question 11:

Find the approximate value of f (5.001), where f (x) = x3 − 7x2 + 15.

Answer:

Let: x=5 x+x=5.001x=0.001fx=x3-7x2+15y=fx=3=125-175+15=-35Now, y=fxdydx=3x2-14x  dy=y=dydxdx=3x2-14x×0.001 =75-70×0.001=0.005 f5.001=y+y=-35+0.005=-34.995

Page No 14.10:

Question 12:

Find the approximate value of log10 1005, given that log10 e = 0.4343.

Answer:

Let: y = fx=log10xHere,  x=1000, x+x=1005x=5dx=x=5For x=1000, y=log101000=log10103=3Now,  y=log10x =logexloge10 dydx=0.4343xdydxx=1000=0.43431000=0.0004343 y=dy=dydxdx=0.0004343×5=0.0021715 log101005=y+y=3.0021715

Page No 14.10:

Question 13:

If the radius of a sphere is measured as 9 cm with an error of 0.03 m, find the approximate error in calculating its surface area.

Answer:

Let x be the radius and y be the surface area of the sphere.

Then, x=9 x=0.03 m=3cmx+x=9+3=12 cmy=4πx2For x=9, y=4π×92=324πdydx=8πxdydxx=9=72π y=dy=dydxdx=72π×3=216π cm2Therefore, the approximate error in the surface area is 216π cm2.Disclaimer: This solution has been created according to the question given in the book. However, the solution given in the book is incorrect.

Page No 14.10:

Question 14:

Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.

Answer:

Let y be the surface area of the cube.

y=6 x2We have xx×100=1Now,dydx=12xy=dy=dydxdx=12x×x100=0.12 x2 m2Hence, approximate change in the surface area of the cube is 0.12x2 m2.

Page No 14.10:

Question 15:

If the radius of a sphere is measured as 7 m with an error of 0.02 m, find the approximate error in calculating its volume.

Answer:

Let x be the radius of the sphere and y be its volume.
y=43πx3Let x be the error in the radius.x=7x=0.02dydx=4πx2dydxx=7=196π y=dy=dydxdx=196π×0.02 =3.92π Hence, the approximate error in calculating the volume of the sphere is 3.92π m3. 

Page No 14.10:

Question 16:

Find the approximate change in the value V of a cube of side x metres caused by increasing the side by 1%.

Answer:


Volume of the cube, V=x3We havex=0.01xdVdx=3x2V=dV=dVdxdx=3x2×0.01x=0.03 x3Hence, the approximate change in the value V of the cube is 0.03x3 m3.Disclaimer: This solution has been created according to the question given in the book. However, the solution in the book is incorrect.



Page No 14.12:

Question 1:

For the function y = x2, if x = 10 and ∆x = 0.1. Find ∆y.

Answer:

y=x2 x=0.1 x=10dydx=2xdydxx=10=20y=dy=dydxdx=20×0.1=2

Page No 14.12:

Question 2:

If y = loge x, then find ∆y when x = 3 and ∆x = 0.03.

Answer:

We have

x=3x=0.03y=logexFor x=3, y=loge3Also, dydx=1xdydxx=3=13y=dy=dydxdx=13 × 0.03=0.01

Page No 14.12:

Question 3:

If the relative error in measuring the radius of a circular plane is α, find the relative error in measuring its area.

Answer:

Let x be the radius and y be the area of the circular plane.
We have xx=α and y=x2.dydx=2xyy=2xydx=2xx2×αx=2αHence, the relative error in the area of the circular plane is 2α.

Page No 14.12:

Question 4:

If the percentage error in the radius of a sphere is α, find the percentage error in its volume.

Answer:

Let V be the volume of the sphere.
V=43πx3We have xx×100=αdVdx=4πx2dVV=4πx2VdxVV=4πx243πx3×xα100VV×100=3αHence, the the percentage error in the volume is 3α.      

Page No 14.12:

Question 5:

A piece of ice is in the form of a cube melts so that the percentage error in the edge of cube is a, then find the percentage error in its volume.

Answer:

Let x be the side and V be the volume of the cube.
V=x3We havexx×100=a dVdx=3x2VV=3x2Vdx=3x2x3×ax100VV×100=3aHence, the percentage error in the volume is 3a.

Page No 14.12:

Question 1:

If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is
(a)1%
(b) 2%
(c) 3%
(d) 4%

Answer:

(a) 1%
Let l be the length if the pendulum and T be the period.

Also, let l be the error in the length and T be the error in the period.We havell×100=2dll×100=2Now, T=2πlgTaking log on both sides, we getlog T=log 2π+12log l-12log gDifferentiating both sides w.r.t. x, we get1TdTdl=12ldTdl=T2ldll×100=2dTT×100dTT×100=22TT×100=1Hence, there is an error of 1% in calculating the period of the pendulum.



Page No 14.13:

Question 2:

If there is an error of a% in measuring the edge of a cube, then percentage error in its surface is
(a) 2a%

(b) a2%

(c) 3a%

(d) none of these

Answer:

(a) 2a%

Let x be the side of the cube and y be its surface area.

xx×100=aAlso, y=6x2dydx=12xyy=12xy×dx =2x×ax100yy×100=2aHence, the error in the surface area is 2a%.

Page No 14.13:

Question 3:

If an error of k% is made in measuring the radius of a sphere, then percentage error in its volume is
(a) k%
(b) 3k%
(c) 2k%
(d) k/3%

Answer:

(b) 3k%
Let x be the radius of the sphere and y be its volume.
Then,
xx×100=kAlso, y=43πx3dydx=4πx2yy=4πx2ydx=4πx243πx3×kx100 yy×100=3kHence, the error in the volume is 3k%.

Page No 14.13:

Question 4:

The height of a cylinder is equal to the radius. If an error of α % is made in the height, then percentage error in its volume is
(a) α %
(b) 2α %
(c) 3α %
(d) none of these

Answer:

(c) 3αα%
Let x be the radius, which is equal to the height of the cylinder. Let y be its volume.

xx×100=αAlso, y=πx2x=πx3         Radius = Height of the cylinderdydx=3πx2yy=3πx2ydx=3x×αx100yy×100=3αHence, the error in the volume of the cylinder is 3α%.

Page No 14.13:

Question 5:

While measuring the side of an equilateral triangle an error of k % is made, the percentage error in its area is
(a) k %
(b) 2k %
(c) k2%
(d) 3k %

Answer:

(b) 2k%

Let x be the side of the triangle and y be its area.

xx×100=kAlso, y=34x2dydx=32xyy=3x2ydx =2x×kx100yy×100=2kHence, the error in the area of the triangle is 2k%.

Page No 14.13:

Question 6:

If loge 4 = 1.3868, then loge 4.01 =
(a) 1.3968
(b) 1.3898
(c) 1.3893
(d) none of these

Answer:

(c) 1.3893
Consider the function y=fx=logex.Let:  x=4x+x=4.01x=0.01For x=4, y=loge4=1.3868y=logexdydx=1xdydxx=4=14y=dy=dydxdx=14× 0.01=0.0025  loge4.01=y+y=1.3893

Page No 14.13:

Question 7:

A sphere of radius 100 mm shrinks to radius 98 mm, then the approximate decrease in its volume is
(a) 12000 π mm3
(b) 800 π mm3
(c) 80000 π mm3
(d) 120 π mm3

Answer:

(c) 80000 π mm3

Let x be the radius of the sphere and y be its volume.

x=100, x+x=98x=-2y=43πx3dydx=4πx2dydxx=100=40000π y=dy=dydxdx=40000π×-2=-80000πHence, the decrease in the volume of the sphere is 80000π mm3.

Page No 14.13:

Question 8:

If the ratio of base radius and height of a cone is 1 : 2 and percentage error in radius is λ %, then the error in its volume is
(a) λ %
(b) 2 λ %
(c) 3 λ %
(d) none of these

Answer:

(c) 3 λ %

Let the radius of the cone be x, the height be 2x and the volume be y.
xx=λ%y=13πx2×2x=23πx3dydx=2πx2yy=2πx2ydx=3x×λxyy=3λ%

Page No 14.13:

Question 9:

The pressure P and volume V of a gas are connected by the relation PV1/4 = constant. The percentage increase in the pressure corresponding to a deminition of 1/2 % in the volume is

(a) 12%

(b) 14%

(c) 18%

(d) none of these

Answer:

(c) 18 %

We have
VV=-12%PV14= constant= k     sayTaking log on both sides, we getlog PV14=log klog P+14log V=log kDifferentiating both sides w.r.t. x, we get1PdPdV+14V=0dPP=-dV4V=-14×-12=18Hence, the increase in the pressure is 18%.

Page No 14.13:

Question 10:

If y = xn, then the ratio of relative errors in y and x is
(a) 1 : 1
(b) 2 : 1
(c) 1 : n
(d) n : 1

Answer:

(d) n:1
Let xx be the relative error in x and yy be the error in y.Now, y=xndydx=n xn-1yy=n xn-1ydxyy=nxn-1xndx =nxxyy:xx=n:1

Page No 14.13:

Question 11:

The approximate value of (33)1/5 is
(a) 2.0125
(b) 2.1
(c) 2.01
(d) none of these

Answer:

(a) 2.0125
Consider the function y= f(x)=x15.
Let: x=32 x+x=33x=1y=x15For x=32, y=2Also, dydx=15x45dydxx=32=180y=dy=dydxdx=180×1=0.0125 3315=y+y=2.0125

Page No 14.13:

Question 12:

The circumference of a circle is measured as 28 cm with an error of 0.01 cm. The percentage error in the area is

(a) 114

(b) 0.01

(c) 17

(d) none of these

Answer:

(a) 114
Let x be the radius of the circle and y be its circumference.
x=28 cmx=0.01 cmx=2πry=πr2=π×x24π2=x24πdydx=x2πyy=x2πydx=2x×0.01yy×100=2x=114Hence, the percentage error in the area is 114.



Page No 14.9:

Question 1:

If y = sin x and x changes from π/2 to 22/14, what is the approximate change in y?

Answer:

Let: x=π2 x+x=2214dx=x=2214-π2=0Now, y=sin xdydx=cos xdydxx=π2= cosπ2=0y=dydxx= 0×0=0y=0

Hence, there is no change in the value of y.

Page No 14.9:

Question 2:

The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume.

Answer:

Let r be the radius of the sphere. r= 10 cmr+r=9.8 cmr=10.0-9.8=0.2 cmVolume of the sphere, V=43πr3dVdr=43π×3r2=4πr2dVdrr=10 cm=4π102=400π cm3/cmChange in the volume of the sphere, V=dVdr×dr=400π×0.2=80π cm3

Page No 14.9:

Question 3:

A circular metal plate expends under heating so that its radius increases by k%. Find the approximate increase in the area of the plate, if the radius of the plate before heating is 10 cm.

Answer:

Let at any time, x be the radius and y be the area of the plate.
Then, y=x2Let x be the change in the radius and y be the change in the area of the plate.  We havexx×100=kWhen x=10, we getx=10k100=k10Now, y=πx2dydx=2πxdydxx=10 cm=20π cm2/cmy=dy=dydxdx=20π×k10=2kπ cm2

Hence, the approximate change in the area of the plate is 2kπ cm2 .

Page No 14.9:

Question 4:

Find the percentage error in calculating the surface area of a cubical box if an error of 1% is made in measuring the lengths of edges of the cube.

Answer:

Let x be the edge of the cube and y be the surface area.

y=x2Let x be the error in x and y be the corresponding error in y.We havexx×100=12x=x100         Let dx=xNow, y=x2dydx=2x y=dydx×x=2x × x100y=2x2100y=2y100yy=2100yy×100=2

Hence, the percentage error in calculating the surface area is 2.

Page No 14.9:

Question 5:

If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere.

Answer:

Let x be the radius and y be the volume of the sphere.
y=43πx3Let x be the error in the radius and y be the error in the volume.Then, xx×100=0.1dxx=11000Now, y=43πx3dydx=4πx2dy=4πx2 dxdyy=4πx2 dx43πx3=3xdxdyy=31000yy×100=0.3

Hence, the percentage error in the calculation of the volume of the sphere is 0.3.

Page No 14.9:

Question 6:

The pressure p and the volume v of a gas are connected by the relation pv1.4 = const. Find the percentage error in p corresponding to a decrease of 1/2% in v.

Answer:

We havepv1.4= constant= k  sayTaking log on both the sides, we getlog pv1.4=log klog p+1.4 log v= log kDifferentiating both the sides w.r.t.  x,  we get1pdpdv+1.4v=0dpp=-1.4 dvvNow, dp=dpdvdv=-1.4pvdvdpp×100=-1.4dvv×100 =-1.4×-12=0.7         Since we are given 12% decrease in vHence, the error in pis 0.7%.

Page No 14.9:

Question 7:

The height of a cone increases by k%, its semi-vertical angle remaining the same. What is the approximate percentage increase (i) in total surface area, and (ii) in the volume, assuming that k is small?

Answer:

Let h be the height, y be the surface area, V be the volume, l be the slant height and r be the radius of the cone.

Let h be the change in the height, r be the change in the radius of base and l be the change in the slant height.Semi-vertical angle ramaining the same. hh=rr=llAlso, hh×100=kThen, hh×100=rr×100=ll×100=k          ...1i Total surface area of the cone, T=πrl+πr2Differentiating both sides w.r.t. r,  we getdTdr=πl+πrdldr+2πrdTdr=πl+πrlr+2πr             From 1, dldr=lr=lr            dTdr=πl+πl+2πr dTdr=2πl+r T=dTdrr=2πl+r×kr100=2krπl+r100  TT×100=2krπl+r1002πrl+r×100=2k %Hence, the percentage increase in total surface area of cone is 2k.ii Volume of cone, V=13πr2hDifferentiating both sides w.r.t. h, we getdVdh=13πr2+13πh2rdrdhdVdh=13πr2+13πh2rrh            From 1, drdh=rh=rhdVdh=13πr2+23πr2dVdh=πr2 V=dVdhdh=πr2×kh100=kπr2h100 VV×100=kπr2h10013πr2h×100=3k %Hence, the percentage increase in the volume of the cone is 3k.

Page No 14.9:

Question 8:

Show that the relative error in computing the volume of a sphere, due to an error in measuring the radius, is approximately equal to three times the relative error in the radius.

Answer:

Let x be the radius of the sphere and y be its volume.
Let x be the error in the radius and V be the approximate error in the volume.y=43πx3dydx=4πx2y=dy=dydxdx=4πx2×xy=3×43πx3×xxy=3×y×xxyy=3xx
Hence proved.

Page No 14.9:

Question 9:

Using differentials, find the approximate values of the following:

(i) 25.02

(ii) 0.00913

(iii) 0.00713

(iv) 401

(v) 1514

(vi) 25514

(vii) 1(2.002)2

(viii) loge 4.04, it being given that log104 = 0.6021 and log10e = 0.4343

(ix) loge 10.02, it being given that loge10 = 2.3026

(x) log10 10.1, it being given that log10e = 0.4343

(xi) cos 61°, it being given that sin60° = 0.86603 and 1° = 0.01745 radian

(xii) 125.1

(xiii) sin2214

(xiv) cos11π36

(xv) 8014

(xvi) 2913

(xvii) 6613

(xviii) 26                  [CBSE 2000]

(xix) 37                    [CBSE 2000]

(xx) 0.48                  [CBSE 2002C]

(xxi) 8214                  [CBSE 2005]

(xxii) 178114

(xxiii) 3315

(xxiv) 36.6

(xxv) 2513

(xxvi) 49.5                 [CBSE 2012]

(xxvii) 3.96832            [CBSE 2014]

(xxviii) 1.9995             [NCERT EXEMPLAR]

(xxix) 0.082               [NCERT EXEMPLAR]

Answer:

(i)
Consider the function y=fx=x.Let: x =25  x+x=25.02Then,x=0.02For x=25, y=25=5Let: dx=x=0.02Now, y=xdydx=12xdydxx=25=110 y=dy=dydxdx=110×0.02=0.002y=0.002 25.02=y+y=5.002

(ii)
Consider the function y=fx=x3.Let: x =0.008x+x=0.009Then, x=0.001For x=0.008, y=0.008=0.2Let: dx=x=0.001Now, y=x3dydx=13x23dydxx=0.008=13×0.04=10.12y=dy=dydxdx=10.12×0.001=1120y=1120=0.008333 0.00913=y+y=0.208333

(iii)
Consider the function y=fx=x.3Let: x =0.008  x+x=0.007Then, x=-0.001For x=0.008, y=0.008=0.2Let: dx=x=-0.001Now, y=x3dydx=13x23dydxx=0.008=13×0.04=10.12 y=dy=dydxdx=10.12×0.001=1120y=1120=0.008333 0.00713=y+y=0.191667

(iv).
Consider the function y=fx=x.Let: x =400 x+x=401Then, x=1For x=400, y=400=20Let: dx=x=1Now, y=xdydx=12xdydxx=400=140 y=dy=dydxdx=140×1=140y=140=0.025 401=y+y=20.025

(v)
Consider the function y=fx=x14.Let: x =16 x+x=15Then, x=-1For x=16, y=1614=2Let: dx=x=-1Now, y=x14dydx=14x34dydxx=16=132 y=dy=dydxdx=132×-1=-132y=-132=-0.03125 1514=y+y=1.96875

(vi)
Consider the function y=fx=x14.Let: x =256x+x=255Then, x=-1For x=256, y=25614=4Let: dx=x=-1Now, y=x14dydx=14x34dydxx=256=1256y=dy=dydxdx=1256×-1=-1256y=-1256=-0.003906 25514=y+y=3.996093.9961

(vii)
Consider the function y=fx=1x2.Let: x =2 x+x=2.002Then, x=-0.002For x=2 , y=122=14Let: dx=x=0.002Now, y=1x2dydx=2x3dydxx=2=14 y=dy=dydxdx=14×-0.002=-0.0005y=-0.0005 12.0022=y+y=0.2495

(viii)
Consider the function y=fx= logex.Let: x = 4 x+x= 4.04Then, x=0.04For x=4, y=loge4=log104log10e=0.60210.4343=1.386368Let: dx=x=0.04Now, y=logexdydx=1xdydxx= 4=14 y=dy=dydxdx=14×0.04=0.01y=0.01 loge4.04=y+y=1.396368

(ix)
Consider the function y=fx=logex.Let: x =10  x+x=10.02Then, x=0.02For x= , y=loge10=2.3026Let: dx=x=0.02Now, y=logexdydx=1xdydxx=10=110 y=dy=dydxdx=110×0.02=0.002y=0.002 loge10.02=y+y=2.3046

(x)
Consider the function y=fx=log10x.Let: x =10 x+x=10.1Then, x=0.1For x= , y=log1010=1Let: dx=x=0.1Now, y=log10x=logexloge10dydx=12.3025xdydxx=10=0.04343 y=dy=dydxdx=0.04343×0.1=0.004343y=0.004343 log1010.1=y+y=1.004343

(xi)
Consider the function y=fx=cos x°.Let: x =60°  x+x=61°Then, x=1°=0.01745For x=60°, y=cos 60°=0.5Let: dx=x=0.01745Now, y=cos xdydx=-sin xdydxx=60=-0.86603 y=dy=dydxdx=-0.86603×0.01745=-0.01511y=-0.01511 cos 61°=y+y=0.484880.48489

(xii)
Consider the function y=fx=1x.Let: x =25 x+x=25.1Then, x=0.1For x= , y=125=0.2Let: dx=x=0.1Now, y=1xdydx=-12x32dydxx=25=-0.004 y=dy=dydxdx=-0.004×0.1=-0.0004y=-0.0004 125.1=y+y=0.1996

(xiii)
Consider the function y=fx=sin x.Let: x =227 x+x=2214Then, x=-2214For x=π, y=sin 227=0Let: dx=x=sin -2214=-sin π2=-1Now, y=sin xdydx=cos xdydxx=227=-1 y=dy=dydxdx=-1×-1=1y=1 sin 2214=y+y=1

(xiv)
Consider the function y=fx=cos x.Let: x =π3 x+x=11π36Then, x=-π36=-5°For x=π3, y=cos π3=0.5Let: dx=x=-sin 5°=-0.08716Now, y=cos xdydx=-sin xdydxx=π3=-0.86603 y=dy=dydxdx=-0.86603×-0.08716=0.075575y=0.075575 cos11π36=y+y=0.5+0.075575=0.575575

(xv)
Consider the function y=fx=x14.Let: x =81 x+x=80Then, x=-1For x=81, y=8114=3Let: dx=x=-1Now, y=x14dydx=14x34dydxx=81=1108 y=dy=dydxdx=1108×-1=-0.009259y=-0.009259 8014=y+y=2.99074

(xvi)
Consider the function y=fx=x13.Let: x =27 x+x=29Then, x=2For x=27, y=2713=3Let: dx=x=2Now, y=x13dydx=13x23dydxx=27=127 y=dy=dydxdx=127×2=0.074y=0.074 2913=y+y=3.074

(xvii)
Consider the function y=fx=x13.Let: x =64 x+x=66Then, x=2For x=64, y=6413=4Let: dx=x=2Now, y=x13dydx=13x23dydxx=64=148 y=dy=dydxdx=148×2=0.042y=0.042 6613=y+y=4.042

(xviii)
Consider the function y=fx=x.Let: x =25x+x=26Then, x=1For x=25, y=25=5Let: dx=x=1Now, y=x1/2dydx=12xdydxx=25=110 y=dy=dydxdx=110×1=0.1y=0.1 26=y+y=5.1

(xix)
Consider the function y=fx=x.Let: x =36 x+x=37Then, x=1For x=36, y=36=6Let: dx=x=1Now, y=x12dydx=12xdydxx=36=112 y=dy=dydxdx=112×1=0.0833y=0.0833 37=y+y=6.0833

(xx)
Consider the function y=fx=x.Let: x =0.49 x+x=0.48Then, x=-0.01For x=0.49,  y=0.49=0.7Let: dx=x=0.01Now, y=x12dydx=12xdydxx=0.49=11.4 y=dy=dydxdx=11.4×-0.01=-0.007143y=-0.007143 0.48=y+y=0.693

(xxi)
Consider the function y=fx=x14.Let: x =81 x+x=82Then, x=1For x=81, y=8114=3Let: dx=x=1Now, y=x14dydx=14x34dydxx=81=1108 y=dy=dydxdx=1108×1=0.009259y=0.009259 8214=y+y=3.009259

(xxii)
Consider the function y=fx=x14.Let: x =1681  x+x=1781Then, x=181For x=1681,  y=168114=23Let: dx=x=181Now, y=x14dydx=14x34dydxx=1681=2732 y=dy=dydxdx=2732×181=196=0.01042y=0.01042 178114=y+y=0.6771

(xxiii)
Consider the function y=fx=x15.Let: x =32 x+x=33Then, x=1For x=33, y=3215=2Let: dx=x=1Now, y=x15dydx=15x45dydxx=32=180 y=dy=dydxdx=180×1=0.0125y=0.0125 3315=y+y=2.0125

(xxiv)
Consider the function y=fx=x.Let: x =36x+x=36.6Then, x=0.6For x=36, y=36=6Let: dx=x=0.6Now, y=x12dydx=12xdydxx=36=112 y=dy=dydxdx=112×0.6=0.05y=0.05 36.6=y+y=6.05

(xv)
Consider the function y=fx=x13.Let: x =27 x+x=25Then, x=-2For x=27, y=2713=3Let: dx=x=-2Now, y=x13dydx=13x23dydxx=27=127 y=dy=dydxdx=127×-2=-0.07407y=-0.07407 2513=y+y=2.9259

(xxvi)
Consider the function y=fx=x.Let: x =49x+x=49.5Then, x=0.5For x=49, y=49=7Let: dx=x=0.5Now, y=x12dydx=12xdydxx=49=114 y=dy=dydxdx=114×0.5=0.0357y=0.0357 49.5=y+y=7.0357

(xxvii)
Consider the function y=fx=x32.Let: x =4  x+x=3.968Then, x=-0.032For x=4,  y=432=8Let: dx=x=-0.032Now, y=x32dydx=3x2dydxx=4=3 y=dy=dydxdx=3×-0.032=-0.096y=-0.096 3.96832=y+y=7.904

(xxviii)
Consider the function y=fx=x5.Let: x=2  x+x=1.999Then, x=-0.001For x=2,  y=25=32Let: dx=x=-0.001Now, y=x5dydx=5x4dydxx=2=80 y=dy=dydxdx=80×-0.001=-0.08y=-0.08 1.9995=y+y=31.92

(xxix)
Consider the function y=fx=x.Let:x=0.0841x+x=0.082Then, x=-0.0021For x=0.0841, y=0.0841=0.29Let: dx=x=-0.0021Now, y=x12dydx=12xdydxx=0.0841=10.58=5029 y=dy=dydxdx=5029×-0.0021=-0.0036y=-0.0036 0.082=y+y=0.2864



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