Question
Find $\frac{\text{dy}}{\text{dx}}$ when x and y are connected by the relation:
$\tan^{-1}(\text{x}^2+\text{y}^2)=\text{a}$

Answer

Consider, $\tan^{-1}(\text{x}^2+\text{y}^2)=\text{a}$
$\text{x}^2+\text{y}^2=\tan\text{a}$
$\Rightarrow\ 2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\ 2\text{y}\cdot\frac{\text{dy}}{\text{dx}}=-2\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-2\text{x}}{2\text{y}}=\frac{-\text{x}}{\text{y}}$

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