Question

If $\cos ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=k$ (a constant), then $\frac{d y}{d x}$ is equal to

• A. $\frac{y}{x}$
• B. $\frac{x}{y}$
• C. $\frac{x^{2}}{y^{2}}$
• D. $\frac{y^{2}}{x^{2}}$

Answer 1

Answer: (A)

Given, $\cos ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=k$
$\Rightarrow \frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\cos k$
On differentiating w.r.t. $x$, we get
$\frac{\left(x^{2}+y^{2}\right)\left(2 x-2 y \frac{d y}{d x}\right)-\left(x^{2}-y^{2}\right)\left(2 x+2 y \frac{d y}{d x}\right)}{\left(x^{2}+y^{2}\right)^{2}} = 0$
$\Rightarrow -4 x^{2} y \frac{d y}{d x}+4 x y^{2}=0$
$\Rightarrow \frac{d y}{d x}=\frac{y}{x}$