Question

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

Answer 1

$y={\sin }^{-1}\left(\frac{2.3}{1+(36{)}^x}\right)+{\sin }^{-1}\left(\frac{2\times 6^x}{1+6^{2}x}\right)$

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

$\cos y=\sqrt{1-\sin y}=\sqrt{1-\frac{{(2\times 6^x)}^2}{{(1+6^{2x})}^2}}$

$=\sqrt{\frac{1+6^{4x}+2\left(6^{2x}\right)-4.6^{2x}}{1+6^{4x}+2\left(6^{2x}\right)}}$

$=\sqrt{\frac{1+6^{4x}+2\left(6^{2x}\right)}{1+6^{4x}+2\left(6^{2x}\right)}}=\sqrt{\frac{{(1-6^{2x})}^2}{{(1+6^{2x})}^2}}$

$\cos y=\frac{(1-6^{2x})}{(1+6^{2x})}$

Differentiation:

$\cos y=\left(\frac{dy}{dx}\right)=2\times \left[\frac{6^x\log 6[1+6^{2x}]-6^{2x}\log 6\left(6^x\right)}{{(1+6^{2x})}^2}\right]$

$\frac{(1-6^{2x})}{(1+6^{2x})}\times \left(\frac{dy}{dx}\right)=2\left[\frac{6^x\log 6+6^{3x}\log 6-26^{3x}\log 6}{{(1+6^{2x})}^x}\right]$

$(1-6^{2x})\frac{dy}{dx}=2\left[\log 6(6^x-6^{3x})\right]$

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

$\frac{dy}{dx}=2\times 6^x\log 6$