Question

Find  \[\frac{dy}{dx}\] in the following case \[\tan^{- 1} \left( x^2 + y^2 \right) = a\] ?

 

Answer 1

\[\text{We have, } \tan^{- 1} \left( x^2 + y^2 \right) = a\]

Differentiate with respect to x, we get,

\[\frac{d}{dx}\left[ \tan^{- 1} \left( x^2 + y^2 \right) \right] = \frac{d}{dx}\left( a \right)\]
\[ \Rightarrow \frac{1}{1 + \left( x^2 + y^2 \right)^2} \times \frac{d}{dx}\left( x^2 + y^2 \right) = 0\]
\[ \Rightarrow \left[ \frac{1}{1 + \left( x^2 + y^2 \right)^2} \right]\left( 2x + 2y\frac{d y}{d x} \right) = 0\]
\[ \Rightarrow 2x + 2y\frac{d y}{d x} = 0\]
\[ \Rightarrow x + y\frac{d y}{d x} = 0\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{x}{y}\]