Question

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

Answer 1

$\frac{d}{dx}{\sin }^{-1}\left[\frac{2^{x+1}{.3}^2}{1+{36}^x}\right]$

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

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

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

$=\frac{({36}^x+1)\left[\ln \right(3){.3}^x{.2}^{x+1}+\ln (2\left)2^{x+1}\frac{d}{dx}\right(x+1\left)3^x\right]-3^x{.2}^{x+1}\left[\ln \right(36){.36}^x+0]}{({36}^x+1{)}^2\sqrt{1-\frac{3^x{.2}^{2(x+1)}}{(1+{36}^x{)}^2}}}$

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

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