Question

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

$\begin{array}{l}\text {  } \frac{x^2-y^2}{x^2+y^2}=\sec ^{-1}\left(e^a\right) \\\frac{\left(x^2+y^2\right)\left(2 x-2 y \frac{d y}{d x}\right)-\left(x^2-y^2\right)\left(2 x+2 y \frac{d y}{d x}\right)}{\left(x^2+y^2\right)^2}=0 \\\Rightarrow \quad x\left(x^2+y^2\right)-y\left(x^2+y^2\right) \frac{d y}{d x} \\=\left(x^2-y^2\right) x+y\left(x^2-y^2\right) \frac{d y}{d x} \\\Rightarrow \quad \quad \quad \frac{d y}{d x}=\frac{2 x y^2}{2 x^2 y}=\frac{y}{x}\end{array}$

So, The correct option will be (B).