Question

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

Answer 1

$y={\sin }^{-1}\left(6x\sqrt{1-9x^2}\right)\frac{-1}{3\sqrt{2}}<x<\frac{1}{3\sqrt{2}}$

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

$\cos y\frac{dy}{dx}=\frac{6x\times -18x}{2\sqrt{1-9x^2}}+6\sqrt{1-9x^2}$

$\cos y\frac{dy}{dx}=\frac{-54x}{\sqrt{1-9x^2}}+6\sqrt{1-9x^2}$

$\frac{dy}{dx}=\frac{6(1-9x^2)-54x}{\sqrt{1-9x^2}}\sec y$

$\frac{dy}{dx}=\frac{6-54x^2-54x}{\sqrt{1-9x^2}}\sec \left({\sin }^{-1}\right(6x\sqrt{1-9x^2}\left)\right)$

$\frac{dy}{dx}=\frac{-5+x^2-54x+6}{\sqrt{1-9x^2}}\times \frac{1}{\cos (\frac{\pi }{2}-{\cos }^{-1}(6x\sqrt{1-9x^2}\left)\right)}$