Q. 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) 0
D) $\frac{\sqrt{1 - y^{2}}}{1 - 2y^{2}}$
Explanation
(b)
Given, $x$ =$\ y\sqrt{1 - y^{2}}$
$\Rightarrow 1 = \frac{dy}{dx}\sqrt{1 - y^{2}} + \frac{y( - y)}{\sqrt{1 - y^{2}}}\frac{dy}{dx}$
$\Rightarrow \ \frac{dy}{dx}\left\lbrack \frac{1 - y^{2} - y^{2}}{\sqrt{1 - y^{2}}} \right\rbrack = 1 \Rightarrow \frac{dy}{dx} = \frac{\sqrt{1 - y^{2}}}{1 - 2y^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.