Find dy dx in the following cases: ^ -1 (x^2+y^2 )=a - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$

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Answer

We have, $\tan^{-1}\big(\text{x}^2+\text{y}^2\big)=\text{a}$Differentiating with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\big[\tan^{-1}\big(\text{x}^2+\text{y}^2\big)\big]=\frac{\text{d}}{\text{dx}}(\text{a})$
$\Rightarrow\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\times\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{y}^2\big)=0$
$\Rightarrow\Big[\frac{1}{1+(\text{x}^2+\text{y}^2)^2}\Big]\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}$

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