Question

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

Answer 1

Given : $si{n}^{-1}\left(\frac{2^{x+1}{.3}^x}{1+{36}^x}\right)$

Let : $y=si{n}^{-1}\left(\frac{2^{x+1}{.3}^x}{1+{36}^x}\right)$

$\Rightarrow y=si{n}^{-1}\left(\frac{{2.2}^x{.3}^x}{1+(6^2{)}^x}\right)$

$=si{n}^{-1}\left(\frac{{2.6}^x}{1+(6^x{)}^2}\right)$

Let $6^x=tan\theta$

$\therefore \Rightarrow si{n}^{-1}\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)$

$=si{n}^{-1}sin2\theta$ $\because sin2\theta =\frac{2tan\theta }{1+ta{n}^2\theta }$

$=2\theta$

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

$\frac{dy}{dx}=2\frac{1}{1+{36}^x}{6}^{x}ln6$

$\therefore \frac{dy}{dx}=\frac{{26}^{x}ln6}{1+{36}^x}$