# Find :

Question:

Find $\frac{d y}{d x}$

$y=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right),-\frac{1}{\sqrt{2}} Solution: The given relationship is$y=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)y=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)\Rightarrow \sin y=2 x \sqrt{1-x^{2}}$Differentiating this relationship with respect to x, we obtain$\cos y \frac{d y}{d x}=2\left[x \frac{d}{d x}\left(\sqrt{1-x^{2}}\right)+\sqrt{1-x^{2}} \frac{d x}{d x}\right]\Rightarrow \sqrt{1-\sin ^{2} y} \frac{d y}{d x}=2\left[\frac{x}{2} \cdot \frac{-2 x}{\sqrt{1-x^{2}}}+\sqrt{1-x^{2}}\right]\Rightarrow \sqrt{1-\left(2 x \sqrt{1-x^{2}}\right)^{2}} \frac{d y}{d x}=2\left[\frac{-x^{2}+1-x^{2}}{\sqrt{1-x^{2}}}\right]\Rightarrow \sqrt{1-4 x^{2}\left(1-x^{2}\right)} \frac{d y}{d x}=2\left[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\right]\Rightarrow \sqrt{\left(1-2 x^{2}\right)^{2}} \frac{d y}{d x}=2\left[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\right]\Rightarrow\left(1-2 x^{2}\right) \frac{d y}{d x}=2\left[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\right]\Rightarrow \frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}}\$