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Question
Find \[\frac{dy}{dx}\] in the following case \[\tan^{- 1} \left( x^2 + y^2 \right) = a\] ?
Solution
\[\text{We have, } \tan^{- 1} \left( x^2 + y^2 \right) = a\]
Differentiate with respect to x, we get,
\[\frac{d}{dx}\left[ \tan^{- 1} \left( x^2 + y^2 \right) \right] = \frac{d}{dx}\left( a \right)\]
\[ \Rightarrow \frac{1}{1 + \left( x^2 + y^2 \right)^2} \times \frac{d}{dx}\left( x^2 + y^2 \right) = 0\]
\[ \Rightarrow \left[ \frac{1}{1 + \left( x^2 + y^2 \right)^2} \right]\left( 2x + 2y\frac{d y}{d x} \right) = 0\]
\[ \Rightarrow 2x + 2y\frac{d y}{d x} = 0\]
\[ \Rightarrow x + y\frac{d y}{d x} = 0\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{x}{y}\]
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