The correct option is D 1/1+x^2 We use chain rule here. ddxtan^ 1(1 x1+x) = (1+x)^2(1+x)^2+(1 x)^2×ddx(1 x1+x) = (1+x)^2(1+x)^2 + (1 x)^2× 2(1+x)^2 ...

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Standard XII

Mathematics

Derivative of Standard Functions

d/dxtan-11-x/...

Question

$\frac{d}{d x} {tan}^{- 1} (\frac{1 - x}{1 + x}) =$ ____________.

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

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\frac{- 1}{1 + x^{2}}

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

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\frac{- 2}{1 + x^{2}}

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Solution

The correct option is D $\frac{- 1}{1 + x^{2}}$
We use chain rule here.
$\frac{d}{d x} {tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{(1 + x)^{2}}{(1 + x)^{2} + (1 - x)^{2}} \times \frac{d}{d x} (\frac{1 - x}{1 + x}) = \frac{(1 + x)^{2}}{(1 + x)^{2} + (1 - x)^{2}} \times \frac{- 2}{(1 + x)^{2}} = - \frac{2}{(1 - x)^{2} + (1 + x)^{2}} = - \frac{1}{1 + x^{2}}$
So option $B$ is the right answer.

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ddxtan−1(1−x1+x)= ____________.

The correct option is D −11+x2We use chain rule here.ddxtan−1(1−x1+x)=(1+x)2(1+x)2+(1−x)2×ddx(1−x1+x)=(1+x)2(1+x)2+(1−x)2×−2(1+x)2=−2(1−x)2+(1+x)2=−11+x2So option B is the right answer.

$\frac{d}{d x} {tan}^{- 1} (\frac{1 - x}{1 + x}) =$ ____________.