Question

Differentiate ${\sin }^{-1}\left[\frac{2^{x+1}{.3}^x}{1+(36{)}^x}\right]$ w.r.t $x$.

Answer 1

We have $y={\sin }^{-1}\left[\frac{2^{x+1}{.3}^x}{1+\left({36}^x\right)}\right]$

$\Rightarrow y={\sin }^{-1}\left[\frac{{2.6}^x}{1+\left(6^{2x}\right)}\right]$

Put $6^x=\tan \theta$

$\Rightarrow \theta ={\tan }^{-1}\left(6^x\right)$

Now,$y={\sin }^{-1}\left[\frac{2\tan \theta }{1+{\tan }^2\theta }\right]$

$\Rightarrow y={\sin }^{-1}\left[\sin \right(2\theta \left)\right]$

$\Rightarrow y=2\theta$

$\Rightarrow y=2{\tan }^{-1}\left(6^x\right)$

We know$\frac{d\left({\tan }^{-1}x\right)}{dx}=\frac{1}{1+x^2}$

Also $\frac{d\left(a^x\right)}{dx}=a^x\ln \left(a\right)$

Therefore, $\frac{dy}{dx}=\frac{2}{1+6^{2x}}\times \frac{d}{dx}\left(6^x\right)$

$\frac{dy}{dx}=\frac{2}{1+6^{2x}}\times 6^x\ln 6$