Question

If $y=si{n}^{-1}\left(6x\sqrt{1-9x^2}\right)$, then $\frac{dy}{dx}$ =?

Answer 1

Formula:

$\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}}$

$y={\sin }^{-1}\left(6x\sqrt{1-9x^2}\right)$

$\frac{dy}{dx}=\frac{1}{\sqrt{1-(6x\sqrt{1-9x^2}{)}^2}}.\frac{d}{dx}\left(6x\sqrt{1-9x^2}\right)$

$=\frac{6}{\sqrt{1-36x^2(1-9x^2)}}[\sqrt{1-9x^2}+x\frac{1}{2\sqrt{1-9x^2}}\frac{d}{dx}(1-9x^2\left)\right]$

$=\frac{6}{\sqrt{1-36x^2+324x^4)}}[\sqrt{1-9x^2}+\frac{x}{2\sqrt{1-9x^2}}(-18x\left)\right]$

$=\frac{6}{\sqrt{(1-18x^2{)}^2}}\left[\frac{1-9x^2-9x^2}{\sqrt{1-9x^2}}\right]$

$=\frac{6}{(1-18x^2)}\left[\frac{1-18x^2}{\sqrt{1-9x^2}}\right]$

$=\frac{6}{\sqrt{1-9x^2}}$