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Question: \(\frac { \mathrm { d } } { \mathrm { dx } }\) \(\left\lbrack \tan^{- 1}\left( \frac{a - x}{1 + ax} ...

ddx\frac { \mathrm { d } } { \mathrm { dx } } [tan⁡−1(a−x1+ax)]\left\lbrack \tan^{- 1}\left( \frac{a - x}{1 + ax} \right) \right\rbrack equals –

A

a1+x2\frac{a}{1 + x^{2}}

B

11+x2\frac{1}{1 + x^{2}}

C

–a1+x2\frac{a}{1 + x^{2}}

D

–11+x2\frac{1}{1 + x^{2}}

Answer

–11+x2\frac{1}{1 + x^{2}}

Explanation

Solution

y = tan–1 a – tan–1 x

Ž dydx\frac{dy}{dx} = −11+x2\frac{- 1}{1 + x^{2}}