Question

If ${\cos }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$ (a constant), then $\frac{dy}{dx}$ 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
$\frac{\left(x^2+y^2\right)\left(2x-2y\frac{dy}{dx}\right)-\left(x^2-y^2\right)\left(2x+2y\frac{dy}{dx}\right)}{{\left(x^2+y^2\right)}^2}=0$
$\Rightarrow -4x^{2}y\frac{dy}{dx}+4x{y}^2=0$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$