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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

If log x is...

Question

If $l o g x$ is the natural logarithm of $x$ to base $e$ , then differentiate the following function with respect to $x$ : $({l o g}^{- 1} x) log x + \sqrt{x} {t a n}^{- 1} x$

\frac{\sqrt{x} + l o g x}{1 + x^{2}} + (\frac{\sqrt{x} + 2}{2 x}) {t a n}^{- 1} x

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{t a n}^{- 1} x (\frac{1}{x} + \frac{1}{2 \sqrt{x}})

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\frac{\sqrt{x}}{x^{2} + 1} + \frac{{t a n}^{- 1} x}{2 \sqrt{x}}

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None of the above

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Solution

The correct option is C $\frac{\sqrt{x}}{x^{2} + 1} + \frac{{t a n}^{- 1} x}{2 \sqrt{x}}$
$\frac{d}{d x} (l n^{- 1} x l n x + \sqrt{x} t a n^{- 1} x)$
$= \frac{d}{d x} ((l n x)^{- 1 + 1} + \sqrt{x} t a n^{- 1} x)$
$= \frac{d}{d x} (1 + \sqrt{x} t a n^{- 1} x)$
$= \frac{d}{d x} (1) + \frac{d}{d x} \sqrt{x} (t a n^{- 1} x) + \frac{d}{d x} t a n^{- 1} x (\sqrt{x})$
$= 0 + \frac{(t a n^{- 1} x)}{2 \sqrt{x}} + \frac{1}{x^{2} + 1} \sqrt{x}$
$\Rightarrow \frac{t a n^{- 1} x}{2 \sqrt{x}} + \frac{\sqrt{x}}{x^{2} + 1}$

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Similar questions

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
If

0 < x < 1

, then

\sqrt{1 + x^{2}} {[{x c o s ({c o t}^{- 1} x) + s i n ({c o t}^{- 1} x)}^{2} - 1]}^{1 / 2}

is equal to
(a)

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

(b)

x

(c)

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

(d)

\sqrt{1 + x^{2}}

equals

A. x tan⁻¹ (x + 1) + C

B. tan^−
1 (x + 1) + C

C. (x + 1) tan⁻¹ x + C

D. tan⁻¹ x + C

Q. Differentiate

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

with respect to

{sin}^{- 1} 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. f(x) = 2x − tan⁻¹ x − log

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

is monotonically increasing when
(a) x > 0
(b) x < 0
(c) x ∈ R
(d) x ∈ R − {0}

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If log x is the natural logarithm of x to base e, then differentiate the following function with respect to x: (log−1x)logx+√xtan−1x

The correct option is C √xx2+1+tan−1x2√xddx(ln−1xlnx+√xtan−1x)=ddx((lnx)−1+1+√xtan−1x)=ddx(1+√xtan−1x)=ddx(1)+ddx√x(tan−1x)+ddxtan−1x(√x)=0+(tan−1x)2√x+1x2+1√x⇒tan−1x2√x+√xx2+1

If $l o g x$ is the natural logarithm of $x$ to base $e$ , then differentiate the following function with respect to $x$ : $({l o g}^{- 1} x) log x + \sqrt{x} {t a n}^{- 1} x$