$\frac{d}{d x} [{tan}^{- 1} [(sec x - tan x)] =$

$\frac{1}{2}$

$- \frac{1}{2}$

$- 1$

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The correct Answer is:C

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Step by step video, text & image solution for (d)/(dx)[tan^(-1)[(secx-tanx)]= by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams.

Updated on:21/07/2023

Class 12 MATHS DIFFERENTIATION

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Knowledge Check

Question 1 - Select One
$\frac{d}{d x} [{tan}^{- 1} {(sec x + {tan}^{2} x}] =$
A $\frac{cos x}{1 + {cos}^{2} x}$
B $\frac{sin x}{1 + {sin}^{2} x}$
C $\frac{sin 2 x}{1 + {sin}^{2} x}$
D $\frac{cos x}{1 + {sin}^{2} x}$
Question 1 - Select One
$\frac{d}{d x} [{tan}^{- 1} (cos e c x + cot x)] =$
A $- 1$
B $- \frac{1}{2}$
C1
D $\frac{1}{2}$
Question 1 - Select One
$\frac{d}{d x} {tan}^{- 1} (sec x + tan x)$ =
A1
B $1 / 2$
C2
DNone of these
Question 1 - Select One
$\frac{d}{d x} [{tan}^{- 1} (1 + x \sqrt{2})] + \frac{d}{d x} [{tan}^{- 1} (1 - x \sqrt{2})] =$
A $\frac{1}{1 - 2 x^{2}}$
B $\frac{- 2}{1 + 2 x^{2}}$
C $\frac{- 2 x}{1 + x^{4}}$
D $\frac{2}{1 + x^{4}}$
Question 1 - Select One
$\frac{d}{d x} [{tan}^{- 1} (\sqrt{x})]$ =
A ${sec}^{- 1} x$
B $\frac{1}{2 \sqrt{x} (1 + x)}$
C $\frac{1}{1 + x}$
D $\frac{2 \sqrt{x}}{1 + x}$
Question 1 - Select One
$\frac{d}{d x} [{tan}^{- 1} (\frac{2 + 3 tan x}{3 - 2 tan x})] =$
A0
B1
C $- 1$
D $3 / 2$
Question 1 - Select One
$\frac{d}{d x} [{tan}^{- 1} (\frac{6 x}{1 + 7 x^{2}})] + \frac{d}{d x} [{tan}^{- 1} (\frac{5 + 2 x}{2 - 5 x})] =$
A $\frac{7}{1 + 49 x^{2}}$
B $\frac{2}{1 + 4 x^{2}}$
C $\frac{- 5}{1 + 25 x^{2}}$
D $\frac{9}{1 + 3 x^{2}}$
Question 1 - Select One
$\frac{d}{d x} [{tan}^{- 1} (\frac{a x - b}{b x + a})] =$
A $\frac{1}{1 + x^{2}} + \frac{a^{2}}{a^{2} + b^{2}}$
B $\frac{- 1}{1 + x^{2}} - \frac{a^{2}}{a^{2} + b^{2}}$
C $\frac{1}{1 + x^{2}} + \frac{a^{2}}{a^{2} + b^{2}}$
Dnone of theses

$\frac{d}{d x} [{tan}^{- 1} [(sec x - tan x)] =$

The correct Answer is:C

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Knowledge Check

$\frac{d}{d x} [{tan}^{- 1} {(sec x + {tan}^{2} x}] =$

$\frac{d}{d x} [{tan}^{- 1} (cos e c x + cot x)] =$

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

$\frac{d}{d x} [{tan}^{- 1} (1 + x \sqrt{2})] + \frac{d}{d x} [{tan}^{- 1} (1 - x \sqrt{2})] =$

$\frac{d}{d x} [{tan}^{- 1} (\sqrt{x})]$ =

$\frac{d}{d x} [{tan}^{- 1} (\frac{2 + 3 tan x}{3 - 2 tan x})] =$

$\frac{d}{d x} [{tan}^{- 1} (\frac{6 x}{1 + 7 x^{2}})] + \frac{d}{d x} [{tan}^{- 1} (\frac{5 + 2 x}{2 - 5 x})] =$

$\frac{d}{d x} [{tan}^{- 1} (\frac{a x - b}{b x + a})] =$

Similar Questions

MARVEL PUBLICATION-DIFFERENTIATION-MULTIPLE CHOICE QUESTIONS (TEST YOUR GRASP - II : CHAPTER 11)

ddx[tan−1[(secx−tanx)]=

The correct Answer is:C

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ddx[tan−1(6x1+7x2)]+ddx[tan−1(5+2x2−5x)]=

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Similar Questions

MARVEL PUBLICATION-DIFFERENTIATION-MULTIPLE CHOICE QUESTIONS (TEST YOUR GRASP - II : CHAPTER 11)

$\frac{d}{d x} [{tan}^{- 1} [(sec x - tan x)] =$

$\frac{d}{d x} [{tan}^{- 1} {(sec x + {tan}^{2} x}] =$

$\frac{d}{d x} [{tan}^{- 1} (cos e c x + cot x)] =$

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

$\frac{d}{d x} [{tan}^{- 1} (1 + x \sqrt{2})] + \frac{d}{d x} [{tan}^{- 1} (1 - x \sqrt{2})] =$

$\frac{d}{d x} [{tan}^{- 1} (\sqrt{x})]$ =

$\frac{d}{d x} [{tan}^{- 1} (\frac{2 + 3 tan x}{3 - 2 tan x})] =$

$\frac{d}{d x} [{tan}^{- 1} (\frac{6 x}{1 + 7 x^{2}})] + \frac{d}{d x} [{tan}^{- 1} (\frac{5 + 2 x}{2 - 5 x})] =$

$\frac{d}{d x} [{tan}^{- 1} (\frac{a x - b}{b x + a})] =$