Question

differentiate tan^-1 x+cot^-1 x with respect x​

Answer 1

Answer:

$y = \tan {}^{ - 1} x + \cot {}^{ - 1} x$

Differnatatiating with respect x

$\frac{dy}{dx} = \frac{d( \tan {}^{ - 1} x)}{dx} + \frac{d( \cot {}^{ - 1} x)}{dx}$

$\frac{dy}{dx} = \frac{1}{1 + x {}^{2} } + ( - \frac{1}{1 + x {}^{2} } )$

$\frac{dy}{dx} = \frac{1}{1 + x {}^{2} } - \frac{1}{1 + x {}^{2} }$

$\frac{dy}{dx} = 0$

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Note :

$\tan {}^{ - 1} x = \frac{1}{1 + x {}^{2} }$

$\cot {}^{ - 1} x = - (\frac{1}{1 + x {}^{2} } )$

The value

$\tan {}^{ - 1} x + \cot {}^{ - 1} x = 1$

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