Question

Differentiate the following with respect to $x$: ${\sin }^{-1}\left(\begin{array}{c}\frac{2^{x+1}{3}^x}{1+(36{)}^x}\end{array}\right).$

Answer 1

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

$y={\sin }^{-1}\left[\frac{2^x+{2.3}^x}{1+(36{)}^x}\right]$

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

We know that, $y=2{\tan }^{-1}{6}^x$

$\frac{dy}{dx}=\frac{2}{1+6^{2x}}{.6}^x\log 6$

$\frac{dy}{dx}=\frac{{2.6}^x\log 6}{1+{36}^x}$