Question

Differentiate the following w.r.t. x :

${\displaystyle {\cos }^{-1} \frac{(1-9^x)}{(1+9^x)}}$

Answer 1

Let y = ${\displaystyle {\cos }^{-1} \frac{(1-9^x)}{(1+9^x)}}$

= ${\displaystyle {\cos }^{-1}\left[\frac{1-{\left(3^x\right)}^2}{1+{\left(3^x\right)}^2}\right]}$
Put 3x = tanθ.
Then θ = tan^–1(3x)

∴ y = ${\displaystyle {\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)}$
= cos–1(cos2θ)
= 2θ
= 2tan–1(3x)
Differentiating w.r.t. x, we get
${\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\left[2{\tan }^{-1}\left(3^x\right)\right]}$

= ${\displaystyle 2\frac{\text{d}}{\text{dx}}\left[{\tan }^{-1}\left(3^x\right)\right]}$

= ${\displaystyle 2\times \frac{1}{1+{\left(3^x\right)}^2}.\frac{\text{d}}{\text{dx}}\left(3^x\right)}$

= ${\displaystyle \frac{2}{1+3^{2x}}\times 3^x\log 3}$

= ${\displaystyle \frac{{2.3}^x\log 3}{1+3^{2x}}}$