Question

If $y={\sin }^{-1}\left(6x\sqrt{1-9x^2}\right),-\frac{1}{3\sqrt{2}}<x<\frac{1}{3\sqrt{2}}$, then find $\frac{dy}{dx}$.

Answer 1

Let $x=\frac{1}{3}\sin \theta$

$\begin{array}{c}y={\sin }^{-1}\left(6.\frac{1}{3}\sin \theta \sqrt{1-9\frac{1}{3^2}{\sin }^2\theta }\right)\\ ={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right)\\ ={\sin }^{-1}\left(2\sin \theta \cos \theta \right)\\ ={\sin }^{-1}\left(\sin 2\theta \right)=2\theta \\ =2{\sin }^{-1}3x\\ \frac{dy}{dx}=2\frac{3}{\sqrt{1-{\left(3x\right)}^2}}=\frac{6}{\sqrt{1-9x^2}}\end{array}$