Question

Find $\frac{dy}{dx}$ if $y={\sin }^{-1}\left[\frac{6x-4\sqrt{1-4x^2}}{5}\right].$

Answer 1

Given:
$y={\sin }^{-1}\left[\frac{6x-4\sqrt{1-4x^2}}{5}\right]$

$={\sin }^{-1}[\frac{6x}{5}-\frac{4}{5}\sqrt{1-(2x{)}^2}]$

$={\sin }^{-1}[(2x.\frac{3}{5})-\frac{4}{5}\sqrt{1-(2x{)}^2}]$

$={\sin }^{-1}[2x.\sqrt{1-{\left(\frac{4}{5}\right)}^2}-\frac{4}{5}.\sqrt{1-(2x{)}^2}]$

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

$[\because {\sin }^{-1}A-{\sin }^{-1}B={\sin }^{-1}(A\sqrt{1-B^2}-B\sqrt{1-A^2}\left)\right]$

$\therefore \frac{dy}{dx}=\frac{1}{\sqrt{1-4x^2}}\times 2-0$

$\frac{dy}{dx}=\frac{2}{\sqrt{1-4x^2}}$