Question

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

Answer 1

Let $u=\frac{2^{x+1}{3}^x}{1+{36}^x}$

$\Rightarrow \frac{du}{dx}=\frac{(1+{36}^x)\frac{d}{dx}\left(2^{x+1}{3}^x\right)-\left(2^{x+1}{3}^x\right)\frac{d}{dx}(1+{36}^x)}{{(1+{36}^x)}^2}$ using quotient rule of differentiation

$=\frac{(1+{36}^x)2\left\{\frac{d}{dx}\left(2^x{3}^x\right)\right\}-2\left(2^x{3}^x\right)\left\{\frac{d}{dx}(1+{36}^x)\right\}}{{(1+{36}^x)}^2}$

$=\frac{(1+{36}^x)2\{2^x{3}^x\log 2+2^x{3}^x\log 3\}-2\left(2^x{3}^x\right){36}^x\log 36}{{(1+{36}^x)}^2}$

$=\frac{2.2^x{3}^x[(1+{36}^x)(\log 2+\log 3)-{36}^x\log 36]}{{(1+{36}^x)}^2}$

$=\frac{2^{x+1}{3}^x[(1+{36}^x)(\log 2+\log 3)-{36}^x\log 36]}{{(1+{36}^x)}^2}$

$=\frac{2^{x+1}{3}^x[\log 6+{36}^x\log 6-2{36}^x\log 6]}{{(1+{36}^x)}^2}$

$=\frac{2^{x+1}{3}^x[\log 6-{36}^x\log 6]}{{(1+{36}^x)}^2}$

$\therefore \frac{du}{dx}=\frac{2^{x+1}{3}^x[1-{36}^x]\log 6}{{(1+{36}^x)}^2}$

$y={\sin }^{-1}u$

$\Rightarrow \frac{dy}{du}=\frac{1}{\sqrt{1-u^2}}$

Using chain rule, we get

$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

$=\frac{1}{\sqrt{1-u^2}}\times \frac{2^{x+1}{3}^x[1-{36}^x]\log 6}{{(1+{36}^x)}^2}$

$=\frac{1}{\sqrt{1-{\left(\frac{2^{x+1}{3}^x[1-{36}^x]\log 6}{{(1+{36}^x)}^2}\right)}^2}}\times \frac{2^{x+1}{3}^x[1-{36}^x]\log 6}{{(1+{36}^x)}^2}$