1. differentiate sin -1 (2x+1.3x)/1+36x.
  2. all in powers.

We have,y = sin-12x+1 . 3x1 + 36x y = sin-12 . 6x1 + 62xput 6x = tan θθ = tan-16xnow,y = sin-12 tan θ1 + tan2θy = sin-1sin 2θy = 2 θ  y = 2 tan-16xdydx= 21 + 62x × ddx6xdydx= 21 + 62x × 6x log 6

  • 182

Hey Hari,

take 6^x=t

sin^(-1) { 2^(x+1).3^x) / 1+36^x

=sin^(-1) { 2^x.2.3^x) / (1+6^(2x))

=sin^(-1) { 2.6^x) / (1+6^(2x))

=sin^(-1) { 2t / 1+t^2}

now take t=tany

=sin^(-1) { 2tan y / 1-(tan^2) y)}

now the quantity within the bracketts is sin2y

so sin^(-1) { sin2y}=2y

=2t

=2(tan^(-1)x)

now differentiate wrt to x

=2/(1+x^2)

=2tan

  • -20

Oops i made a mistake

t=tan^(-1)(6^x)

So,

we have to find

2d(tan^(-1)(6^x))/dx

=2/(1+6^(2x)).d(6^x)/dx

=(2/1+36^x).6^x.log6

=2.2^x.3^x/1+36^x

=2^(x+1).3^x/1+36^x

  • -13
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