Question

Differentiate \[\sin^{- 1} \left\{ \frac{2^{x + 1} \cdot 3^x}{1 + \left(36 \right)^x} \right\}\] with respect to x.

Answer 1

We have, 

y = `sin^-1[(2^(x + 1) .  3^x)/(1 + 36^x)]`

y = `sin^-1[(2 . 6^x)/(1 + 6^(2x))]`

Put 6^x = tan θ

⇒ θ = tan^–1(6^x)

Now, 

y = `sin^-1[(2  tanθ)/(1 + tan^2θ)]`

y = sin^–1 [sin 2θ]

y = 2θ

y = 2tan^–1(6^x)

`dy/dx = 2/(1 + 6^(2x)) xx d/dx (6^x)`

`dy/dx = 2/(1 + 6^(2x)) xx 6^x log 6`

`dy/dx = 2/(1 + (36)^(x)) xx 6^x log 6`