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

We have,

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

$$\begin{gathered} \mathrm{So},\frac{dy}{dx}=\frac{d}{dx}\left[{\sin }^{-1}\left(6x\sqrt{1-9x^2}\right)\right] \\ =\frac{d}{dx}\left[{\sin }^{-1}\left(6x\sqrt{1-9x^2}\right)\right] \\ =\frac{1}{\sqrt{1-{\left(6x\sqrt{1-9x^2}\right)}^2}}\times \frac{d}{dx}\left(6x\sqrt{1-9x^2}\right) \\ =\frac{1}{\sqrt{1-\left[36x^2\left(1-9x^2\right)\right]}}\times \left(6x\frac{d}{dx}\sqrt{1-9x^2}+\sqrt{1-9x^2}\frac{d}{dx}\left(6x\right)\right) \\ =\frac{1}{\sqrt{1-36x^2-324x^4}}\times \left(6x\times \frac{1}{2\sqrt{1-9x^2}}\frac{d}{dx}\left(1-9x^2\right)+\sqrt{1-9x^2}\left(6\right)\right) \\ =\frac{1}{\sqrt{1-36x^2-324x^4}}\times \left(6x\times \frac{1}{2\sqrt{1-9x^2}}\times \left(-18x\right)+6\sqrt{1-9x^2}\right) \\ =\frac{1}{\sqrt{1-36x^2-324x^4}}\times \left(\frac{-54x^2}{\sqrt{1-9x^2}}+6\sqrt{1-9x^2}\right) \\ =\frac{1}{\sqrt{1-36x^2-324x^4}}\times \left(\frac{-54x^2+6\left(1-9x^2\right)}{\sqrt{1-9x^2}}\right) \\ =\frac{-54x^2+6-54x^2}{\sqrt{1-9x^2}\sqrt{1-36x^2-324x^4}} \\ =\frac{6-108x^2}{\sqrt{1-9x^2}\sqrt{1-36x^2-324x^4}} \end{gathered}$$