Derivative of tan -1x-1-tan -1x w-r-t- tan -1x is equal to

The correct option is A 1/(1+tan ^ 1x)^2 nbsp; d / dx ( tan ^ 1 x nbsp;/ 1+tan ^ 1 x nbsp; nbsp;) nbsp;=( 1+tan ^ 1 x nbsp;) 1 ...

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Standard XII

Mathematics

Parametric Differentiation

Derivative of...

Question

Derivative of $\frac{{tan}^{- 1} x}{1 + {tan}^{- 1} x}$ w.r.t. ${tan}^{- 1} x$ is equal to

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

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

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

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log (1 + {tan}^{- 1} x)

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Solution

The correct option is A $\frac{1}{(1 + {tan}^{- 1} x)^{2}}$
$\frac{d}{d x} (\frac{{tan}^{- 1} x}{1 + {tan}^{- 1} x})$
$= \frac{(1 + {tan}^{- 1} x) \frac{1}{1 + x^{2}} - ({tan}^{- 1} x) \frac{1}{1 + x^{2}}}{{(1 + {tan}^{- 1} x)}^{2}}$
$= \frac{1}{{(1 + x^{2}) (1 + {tan}^{- 1} x)}^{2}}$
$\frac{d}{d x} ({tan}^{- 1} x) = \frac{1}{1 + x^{2}}$
Therefore derivative of $\frac{{tan}^{- 1} x}{1 + {tan}^{- 1} x}$ w.r.t ${tan}^{- 1} x$ is
$= \frac{1}{(1 + x^{2}) {(1 + {tan}^{- 1} x)}^{2}} \times \frac{(1 + x^{2})}{1} = \frac{1}{{(1 + {tan}^{- 1} x)}^{2}}$

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Similar questions

Q. Assertion (

A

): Derivative of

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

with respect to

{tan}^{- 1} x

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

Reason (

R

): Derivative of

\frac{x}{1 + x}

with respect to

x

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

Solve the following equations for x:
(i) tan⁻¹2x + tan⁻¹3x = nπ + $\frac{3 π}{4}$

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹ $\frac{8}{31}$

(iii) $\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}$

(iv) sin⁻¹x + sin⁻¹2x = $\frac{π}{3}$

(v) $3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}$

(vi) $\cos^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6}$

(vii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x

(viii) tan (cos⁻¹x) = sin $(\cot^{- 1} \frac{1}{2})$

(ix) tan⁻¹ $(\frac{1 - x}{1 + x}) - \frac{1}{2}$ tan⁻¹x = 0, where x > 0

(x) cot⁻¹x − cot⁻¹(x + 2) = $\frac{π}{12}$ , x > 0

(xi) $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0$

(xii) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹ $(\frac{8}{79})$ , x > 0

(xiii) $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$

\int_{- 1}^{3} (tan^{- 1} \frac{x}{x^{2} + 1} + tan^{- 1} \frac{x^{2} + 1}{x}) d x

Q. Assertion :

\int_{- 1}^{3} {{tan}^{- 1} (\frac{x}{1 + x^{2}}) + {tan}^{- 1} (\frac{x^{2} + 1}{x})} d x = 2 π

Reason:

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

and

{tan}^{- 1} t = {cot}^{- 1} \frac{1}{t}

\int_{- 1}^{3} (T a n^{- 1} \frac{x}{x^{2} + 1} + T a n^{- 1} \frac{x^{2} + 1}{x}) d x =

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Derivative of tan−1x1+tan−1x w.r.t. tan−1x is equal to

The correct option is A 1(1+tan−1x)2ddx(tan−1x1+tan−1x)=(1+tan−1x)11+x2−(tan−1x)11+x2(1+tan−1x)2=1(1+x2)(1+tan−1x)2ddx(tan−1x)=11+x2Therefore derivative of tan−1x1+tan−1x w.r.t tan−1x is=1(1+x2)(1+tan−1x)2×(1+x2)1=1(1+tan−1x)2

Derivative of $\frac{{tan}^{- 1} x}{1 + {tan}^{- 1} x}$ w.r.t. ${tan}^{- 1} x$ is equal to