Question

If $y=b{\tan }^{-1}(\frac{x}{a}+{\tan }^{-1}\frac{y}{x})$, find $\frac{dy}{dx}$.

Answer 1

$y=b{\tan }^{-1}(\frac{x}{a}+{\tan }^{-1}\frac{y}{x})$

$\frac{y}{b}={\tan }^{-1}(\frac{x}{a}+{\tan }^{-1}\frac{y}{x})$

$\tan \frac{y}{b}=\frac{x}{a}+{\tan }^{-1}\frac{y}{x}$

Differentiating both sides w.r.t. x, we get,

$\frac{1}{b}{\sec }^2\left(\frac{y}{b}\right)\frac{dy}{dx}=\frac{1}{a}+\frac{1}{1+(\frac{y}{x}{)}^2}\times \frac{x\frac{dy}{dx}-y}{x^2}$

$\frac{1}{b}{\sec }^2\left(\frac{y}{b}\right)\frac{dy}{dx}=\frac{1}{a}+\frac{x\frac{dy}{dx}-y}{x^2+y^2}$

$\frac{dy}{dx}\left[\frac{1}{b}{\sec }^2\right(\frac{y}{b})-\frac{x}{x^2+y^2}]=\frac{1}{a}-\frac{y}{x^2+y^2}$

Therefore,

$\frac{dy}{dx}=\frac{\frac{1}{a}-\frac{y}{x^2+y^2}}{\frac{1}{b}{\sec }^2\left(\frac{y}{b}\right)-\frac{x}{x^2+y^2}}$