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Question: Differentiation of \(\left( \frac{\tan^{–1}x}{1 + \tan^{–1}x} \right)\)w.r.t (tan<sup>–1</sup>x) is ...

Differentiation of (tan⁡–1x1+tan⁡–1x)\left( \frac{\tan^{–1}x}{1 + \tan^{–1}x} \right)w.r.t (tan–1x) is –

A

1

B

–1

C

1(1+tan⁡–1x)2\frac{1}{(1 + \tan^{–1}x)^{2}}

D

0

Answer

1(1+tan⁡–1x)2\frac{1}{(1 + \tan^{–1}x)^{2}}

Explanation

Solution

y = tan⁡–11+tan⁡–1x\frac{\tan^{–1}}{1 + \tan^{–1}x} & z = tan–1x

=1y =z1+z\frac{z}{1 + z}Ždydz\frac{dy}{dz}=(1+z).1–z.1(i+z)2\frac{(1 + z).1–z.1}{(i + z)^{2}}=1(1+z)2\frac{1}{(1 + z)^{2}}

= 1(1+tan⁡–1x)2\frac{1}{(1 + \tan^{–1}x)^{2}}