Question

Find $\frac{dy}{dx}$ for $y={\sin }^{-1}\left(\frac{6x-4\sqrt{1-4x^2}}{5}\right)$

Answer 1

Consider the given equation,

$y={\sin }^{-1}\left(\frac{6x-4\sqrt{1-4x^2}}{5}\right)$

$\sin y=\frac{6x-4\sqrt{1-4x^2}}{5}$

$5\sin y=6x-4\sqrt{1-4x^2}$

Differentiate both sides with respect to x,

$5\frac{d}{dx}\sin y=\frac{d}{dx}(6x-4\sqrt{1-4x^2})$

$5\cos y\frac{dy}{dx}=6.1-4.\frac{1}{2\sqrt{1-4x^2}}\frac{d}{dx}(1-4x^2)$

$5\cos y\frac{dy}{dx}=6-\frac{2}{\sqrt{1-4x^2}}.(0-8x)$

$5\cos y\frac{dy}{dx}=\frac{6\sqrt{1-4x^2}+16x}{\sqrt{1-4x^2}}$

$\frac{dy}{dx}={\cos }^{-1}\left(\frac{6\sqrt{1-4x^2}+16x}{5\sqrt{1-4x^2}}\right)$

Hence, this is the answer.