Question

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

• A. $\frac{\frac{1}{a}-\frac{y}{x^2+y^2}}{\frac{1}{b}{\sec }^2\left(\frac{y}{b}\right)-\frac{x}{x^2+y^2}}$
• B. $\frac{\frac{1}{a}+\frac{y}{x^2+y^2}}{\frac{1}{b}{\sec }^2\left(\frac{y}{b}\right)+\frac{x}{x^2+y^2}}$
• C. $\frac{\frac{1}{a}-\frac{y}{x^2+y^2}}{\frac{1}{b}{\sec }^2\left(\frac{y}{b}\right)-\frac{x^2+y^2}{x}}$
• D. None of these

Answer 1

The correct option is A $\frac{\frac{1}{a}-\frac{y}{x^2+y^2}}{\frac{1}{b}{\sec }^2\left(\frac{y}{b}\right)-\frac{x}{x^2+y^2}}$

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

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

Differentiating both sides w.r.t. $x$, we get
$\frac{1}{b}{\sec }^2\left(\frac{y}{b}\right)\frac{dy}{dx}=\frac{1}{a}+\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\times \frac{x\frac{dy}{dx}-y}{x^2}$
or $\frac{1}{b}{\sec }^2\left(\frac{y}{b}\right)\frac{dy}{dx}=\frac{1}{a}+\frac{x\frac{dy}{dx}-y}{x^2+y^2}$

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

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