D - dx tan -11- x -1- x -A- -2-1 1- x 2B- 2-1- x 2- 1-1- x 2D- -1-1- x 2

The correct option is D 11+x^2ddxtan^ 1(1 x1+x) =11+(1 x1+x)^2×(1+x)( 1) (1 x)(1)(1+x)^2 = 2(1+x)^2+(1 x)^2 = 22(1+x^2)= 11+x^2 ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

d / dx tan -1...

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{2}{1 + x^{2}}

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

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Solution

The correct option is B $\frac{- 1}{1 + x^{2}}$
$\frac{d}{d x} {tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{1 + {(\frac{1 - x}{1 + x})}^{2}} \times \frac{(1 + x) (- 1) - (1 - x) (1)}{(1 + x)^{2}}$
$= \frac{- 2}{(1 + x)^{2} + (1 - x)^{2}} = \frac{- 2}{2 (1 + x^{2})} = \frac{- 1}{1 + x^{2}}$

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Similar questions

If $y = s i n^{- 1} (x \sqrt{1 - x} + \sqrt{x} \sqrt{1 - x^{2}})$ then $\frac{d y}{d x} =$

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

____________.

$s i n (t a n^{- 1} x), | x | < 1$ , is equal to
a) $\frac{x}{\sqrt{1 - x^{2}}}$
b) $\frac{1}{\sqrt{1 - x^{2}}}$
c) $\frac{1}{\sqrt{1 + x^{2}}}$
d) $\frac{x}{\sqrt{1 + x^{2}}}$

$I f f (x) = x - x^{2} + x^{3} - x^{4} + . . . . . . . \infty, | x | < 1, t h e n f^{'} (x) =$

Q. If

y = s i n^{- 1} \frac{\sqrt{(1 + x)} + \sqrt{(1 - x)}}{2}

, then

\frac{d y}{d x} =

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ddxtan−1(1−x1+x)=

The correct option is B −11+x2ddxtan−1(1−x1+x)=11+(1−x1+x)2×(1+x)(−1)−(1−x)(1)(1+x)2 =−2(1+x)2+(1−x)2=−22(1+x2)=−11+x2

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

The correct option is B $\frac{- 1}{1 + x^{2}}$
$\frac{d}{d x} {tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{1 + {(\frac{1 - x}{1 + x})}^{2}} \times \frac{(1 + x) (- 1) - (1 - x) (1)}{(1 + x)^{2}}$
$= \frac{- 2}{(1 + x)^{2} + (1 - x)^{2}} = \frac{- 2}{2 (1 + x^{2})} = \frac{- 1}{1 + x^{2}}$