Question

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

Answer 1

Let

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

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

put $6^x=tan\theta$

$\theta =ta{n}^{-1}{6}^x$

$=si{n}^{-1}\left[\frac{2tan\theta }{1+ta{n}^2\theta }\right]$

$=si{n}^{-1}\left(sin2\theta \right)$

$=2\theta$

$y=2ta{n}^{-1}{6}^x$

$\frac{dy}{dx}=2\times \frac{1}{1+(6^x{)}^2}\times 6^x\log 6=\frac{{2.6}^x\log 6}{1+6^{2x}}$