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Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

If gx=xtan ...

Question

If $g (x) = x {tan}^{- 1} x$ then the value of $g^{'} (1)$ equals-

\frac{1}{2}

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\frac{π}{4}

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\frac{1}{2} - \frac{π}{4}

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\frac{1}{2} + \frac{π}{4}

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Solution

The correct option is D $\frac{1}{2} + \frac{π}{4}$
$g (x) = x {tan}^{- 1} x$
$\Rightarrow g^{'} (x) = {tan}^{- 1} x + \frac{x}{1 + x^{2}}$
$\Rightarrow g^{'} (1) = {tan}^{- 1} (1) + \frac{1}{1 + {(1)}^{2}} = \frac{π}{4} + \frac{1}{2}$

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Similar questions

Q. The value of

lim x \to 0 \frac{1}{x} ({tan}^{- 1} (\frac{x + 1}{2 x + 1}) - \frac{π}{4})

is equal to.

Prove that: $t a n^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}) = \frac{π}{4} - \frac{1}{2} c o s^{- 1} x; - \frac{1}{\sqrt{2}} \leq x \leq 1$ .

OR If $t a n^{- 1} (\frac{x - 2}{x - 4}) + t a n^{- 1} (\frac{x + 2}{x + 4}) = \frac{π}{4}$ , find the value of x.

Q. If

{tan}^{- 1} \frac{x - 1}{x - 2} + {tan}^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}

, then find the value of

x

Q. If

f (x) = s i n x + c o s x, g (x) = x^{2} - 1

, then g{f(x)} is invertible in the domain

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

.
(a)

{s i n}^{- 1} (3 x / 5) + {s i n}^{- 1} (4 x / 5) = {s i n}^{- 1} x

(b)

{c o s}^{- 1} x + {s i n}^{- 1} (\frac{1}{2} x) = \frac{π}{6}

a \leq {t a n}^{- 1} (\frac{1 - x}{1 + x}) \leq b

where

0 \leq x \leq 1

then

(a, b) =

(a)

(0, π)

(b)

(0, π / 4)

(c)

(- π / 4, π / 4)

(d)

(π / 4, π / 2)

(d) If

a \leq {({s i n}^{- 1} x)}^{3} + {({c o s}^{- 1} x)}^{3} \leq b

then (a,b) is equal to

(\frac{π^{3}}{32}, \frac{7 π^{3}}{8})

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If g(x)=xtan−1x then the value of g′(1) equals-

The correct option is D 12+π4g(x)=xtan−1x⇒g′(x)=tan−1x+x1+x2⇒g′(1)=tan−1(1)+11+(1)2=π4+12

If $g (x) = x {tan}^{- 1} x$ then the value of $g^{'} (1)$ equals-

The correct option is D $\frac{1}{2} + \frac{π}{4}$
$g (x) = x {tan}^{- 1} x$
$\Rightarrow g^{'} (x) = {tan}^{- 1} x + \frac{x}{1 + x^{2}}$
$\Rightarrow g^{'} (1) = {tan}^{- 1} (1) + \frac{1}{1 + {(1)}^{2}} = \frac{π}{4} + \frac{1}{2}$