Question

If $y={sin}^{-1}\left[\frac{5x+12\sqrt{1-x^2}}{13}\right]$ then find $\frac{dy}{dx}$

Answer 1

$y={\sin }^{-1}\left[\frac{5x+12\sqrt{1-x^2}}{13}\right]$

$\frac{dy}{dx}=\frac{1}{\sqrt{1-{\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)}^2}}\frac{d}{dx}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$

$=\frac{13}{\sqrt{169-{(5x+12\sqrt{1-x^2})}^2}}\times \frac{1}{13}[5+12\frac{d}{dx}\left(\sqrt{1-x^2}\right)]$

$=\frac{1}{\sqrt{169-25x^2-12(1-x^2)-120x\sqrt{1-x^2}}}[5-\frac{24x}{2\sqrt{1-x^2}}]$

$=\frac{1}{\sqrt{169-25x^2-12+12x^2-120x\sqrt{1-x^2}}}[5-\frac{12x}{\sqrt{1-x^2}}]$

$=\frac{1}{\sqrt{154-13x^2-120x\sqrt{1-x^2}}}\left[\frac{5\sqrt{1-x^2}-12x}{\sqrt{1-x^2}}\right]$

$=\frac{5\sqrt{1-x^2}-12x}{\left(\sqrt{1-x^2}\right)\sqrt{154-13x^2-120x\sqrt{1-x^2}}}$