Question

If $ x=y\sqrt{1-y^{2}} $ , then $ \frac{dy}{dx} $ is equal to

• A. $ x $
• B. $ \frac{\sqrt{1-y^{2}}}{1+2y^{2}} $
• C. $ \frac{\sqrt{1-y^{2}}}{1-2y^{2}} $
• D. $ 0 $

Answer 1

Answer: (C)

We have $x=y \sqrt{1-y^{2}}$
On differentiating both sides w.r.t. $x$, we get
$ 1=\frac{d y}{d x} \sqrt{1-y^{2}}+\frac{y(-y)}{\sqrt{1-y^{2}}} \frac{d y}{d x} $
$ \Rightarrow \, \frac{d y}{d x}\left[\sqrt{1-y^{2}}-\frac{y^{2}}{\sqrt{1-y^{2}}}\right]=1$
$ \Rightarrow \, \frac{d y}{d x}\left[\frac{1-y^{2}-y^{2}}{\sqrt{1-y^{2}}}\right]=1$
$ \Rightarrow \, \frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{1-2 y^{2}} $