Q. If $\sec\left( \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \right) = e^{a},then\frac{dy}{dx\ }$ is equal to

A) $\frac{y^{2}}{x^{2}}$
B) $\frac{y}{x}$
C) $\frac{x}{y}$
D) $\frac{x^{2} - y^{2}}{x^{2} + y^{2}}$
Explanation

(b)
Given, $\left( \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \right) = {\sec^{- 1}e}^{a}$
$\Rightarrow$ $\frac{\begin{bmatrix}
\left( x^{2} + y^{2} \right)\left( 2x - 2y\frac{dy}{dx} \right) \\
–\left( x^{2} - y^{2} \right)\left( 2x + 2y\frac{dy}{dx} \right)
\end{bmatrix}}{\left( x^{2} + y^{2} \right)^{2}}$ =0
$\Rightarrow \left( 2x^{3} + 2xy^{2} - 2x^{3} + 2xy^{2} \right) - 2y\frac{dy}{dx}$
$\left( x^{2} + y^{2} + x^{2} - y^{2} \right) = 0$
$\Rightarrow 4xy^{2} - 4x^{2}y\ \frac{dy}{dx} = 0$
$\Rightarrow \ \frac{dy}{dx} = \frac{y}{x}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.