Diffferentiate tan -1x-1-tan -1x w-r-t- tan-1x-

The correct option is A 1/ ( 1+tan ^ 1x )^2. Let nbsp; y=tan ^ 1x/1+tan ^ 1x nbsp; and nbsp; z=tan^ 1x. ∴ y= z/1+z= z+1 1/1+z= 1 1 ...

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Standard XII

Mathematics

General Solution of Trigonometric Equation

Diffferentiat...

Question

Diffferentiate $\frac{{tan}^{- 1} x}{1 + {tan}^{- 1} x}$ w.r.t. ${tan}^{- 1} x .$

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

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\frac{1}{{(1 - {tan}^{- 1} x)}^{4}} .

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\frac{1}{{(1 + {tan}^{- 1} x)}^{4}} .

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

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Solution

The correct option is A $\frac{1}{{(1 + {tan}^{- 1} x)}^{2}} .$
Let $y = \frac{{tan}^{- 1} x}{1 + {tan}^{- 1} x}$ and $z = {tan}^{- 1} x .$
$∴ y = \frac{z}{1 + z} = \frac{z + 1 - 1}{1 + z} = 1 - \frac{1}{1 + z} .$
$∴ \frac{d y}{d z} = \frac{1}{{(1 + z)}^{2}} = \frac{1}{{(1 + {tan}^{- 1} x)}^{2}} .$

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Similar questions

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

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

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}

Q. Differentiate

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

w.r.t.

{tan}^{- 1} x .

Q. The value of integral

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

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Diffferentiate tan−1x1+tan−1x w.r.t. tan−1x.

The correct option is A 1(1+tan−1x)2.Let y=tan−1x1+tan−1x and z=tan−1x.∴y=z1+z=z+1−11+z=1−11+z.∴dydz=1(1+z)2=1(1+tan−1x)2.

Diffferentiate $\frac{{tan}^{- 1} x}{1 + {tan}^{- 1} x}$ w.r.t. ${tan}^{- 1} x .$