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

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Given, } \cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}ight)=k \ & \Rightarrow \quad \frac{x^2-y^2}{x^2+y^2}=\cos k \end{aligned}$ On differentiating w.r.t. $x$, we get $\begin{array}{r}\frac{\left(x^2+y^2ight)\left(2 x-2 y \frac{d y}{d x}ight)-\left(x^2-y^2ight)\left(2 x+2 y \frac{d y}{d x}ight)}{\left(x^2+y^2ight)^2} \ =0\end{array}$ $\begin{aligned} \Rightarrow & -4 x^2 y \frac{d y}{d x}+4 x y^2 =0 \ \Rightarrow & \frac{d y}{d x} =\frac{y}{x}\end{aligned}$

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

Solution