Question

Find $\frac{dy}{dx}$ in each of the following cases:

${\tan }^{-1}\left(x^2+y^2\right)=a$

Answer 1

$\mathrm{We}\mathrm{have},{\tan }^{-1}\left(x^2+y^2\right)=a$
Differentiate with respect to x, we get,
$$\begin{gathered} \frac{d}{dx}\left[{\tan }^{-1}\left(x^2+y^2\right)\right]=\frac{d}{dx}\left(a\right) \\ \Rightarrow \frac{1}{1+{\left(x^2+y^2\right)}^2}\times \frac{d}{dx}\left(x^2+y^2\right)=0 \\ \Rightarrow \left[\frac{1}{1+{\left(x^2+y^2\right)}^2}\right]\left(2x+2y\frac{\mathit{d}y}{\mathit{d}x}\right)=0 \\ \Rightarrow 2x+2y\frac{\mathit{d}y}{\mathit{d}x}=0 \\ \Rightarrow x+y\frac{\mathit{d}y}{\mathit{d}x}=0 \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=-\frac{x}{y} \end{gathered}$$