Question

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. $\frac{x^2-y^2}{x^2+y^2}$

Answer 1

The correct option is B

$\frac{y}{x}$

Explanation for the correct option:

Step 1: Using componendo dividendo rule, we get

Given, $\sec \left(\frac{x^2-y^2}{x^2+y^2}\right)=e^a$.

$\frac{x^2-y^2}{x^2+y^2}={\sec }^{-1}{e}^a$
Here apply componendo dividendo rule, we get
$$\begin{gathered} \frac{2x^2}{-2y^2}=\frac{\left({\sec }^{-1}{e}^a+1\right)}{\left({\sec }^{-1}{e}^a-1\right)} \\ \frac{y^2}{x^2}=\frac{\left(1-{\sec }^{-1}{e}^a\right)}{\left(1+{\sec }^{-1}{e}^a\right)}\dots \left(1\right) \\ {y}^2=x^2\frac{\left(1-{\sec }^{-1}{e}^a\right)}{\left(1+{\sec }^{-1}{e}^a\right)} \\ 2y\frac{dy}{dx}=2x\frac{\left(1-{\sec }^{-1}{e}^a\right)}{\left(1+{\sec }^{-1}{e}^a\right)} \end{gathered}$$

Step 2: Finding the value of $\frac{dy}{dx}$

$\begin{array}{rcl}\therefore \frac{dy}{dx}& =& \frac{x}{y}\frac{\left(1-{\sec }^{-1}{e}^a\right)}{\left(1+{\sec }^{-1}{e}^a\right)}\\ & =& \frac{x}{y}\times \frac{y^2}{x^2}(\text{from (1))}\\ & =& \frac{y}{x}\end{array}$
Hence, option B is correct.