Download App

STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
${d \over {dx}}{\tan ^{ - 1}}{{4\sqrt x } \over {1 - 4x}} = $
  • A
    ${1 \over {\sqrt x (1 + 4x)}}$
  • ✓
    ${2 \over {\sqrt x (1 + 4x)}}$
  • C
    ${4 \over {\sqrt x (1 + 4x)}}$
  • D
    None of these
✓

Answer

Correct option: B.
${2 \over {\sqrt x (1 + 4x)}}$
b
(b) $\frac{d}{{dx}}{\tan ^{ - 1}}\frac{{4\sqrt x }}{{1 - 4x}}$

$ = \frac{1}{{1 + {{\left( {\frac{{4\sqrt x }}{{1 - 4x}}} \right)}^2}}}.\left[ {\frac{{(1 - 4x)4\left( {\frac{1}{{2\sqrt x }}} \right) - 4\sqrt x ( - 4)}}{{{{(1 - 4x)}^2}}}} \right]$

$ = \frac{{2(1 + 4x)}}{{\sqrt x {{(1 + 4x)}^2}}} = \frac{2}{{\sqrt x (1 + 4x)}}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Three coins are tossed. If one of them shows tail, then the probability that all three coins show tail, is
→
$\int_0^{\pi /4} {} (\cos x - \sin x)dx + \int_{\pi /4}^{5\pi /4} {} (\sin x - \cos x)dx$ $ + \int_{2\pi }^{\pi /4} {} (\cos x - \sin x)\,dx$ is equal to
→
The nth derivative of $x{e^x}$ vanishes when
→
$\int_{ - \pi /2}^{\pi /2} {{{\sin }^2}x{{\cos }^2}x(\sin x + \cos x)\,dx = } $
→
A unit vector perpendicular to the plane determined by the points  $ (1, -1, 2), (2, 0, -1) $ and  $ (0, 2, 1) $ is
→
Real part of ${e^{{e^{i\theta }}}}$is
→
Given vertices $A(1,\,1),B(4,\, - 2)$ and $C(5,\,5)$ of a triangle, then the equation of the perpendicular dropped from $C$ to the interior bisector of the angle $A$ is
→
The equation of the line of shortest distance of the lines $\frac{{x - 6}}{3} = \frac{{y - 7}}{{ - 1}} = \frac{{z - 4}}{1}$ and $\frac{x}{{ - 3}} = \frac{{y + 9}}{2} = \frac{{z - 2}}{4}$
→
Let $\overrightarrow{ a }=\hat{ i }+2 \hat{ j }+ k , \overrightarrow{ b }=3(\hat{ i }-\hat{ j }+ k )$. Let $\overrightarrow{ c }$ be the vector such that $\overrightarrow{ a } \times \overrightarrow{ c }=\overrightarrow{ b }$ and $\overrightarrow{ a } \cdot \overrightarrow{ c }=3$. Then $\overrightarrow{ a } \cdot((\overrightarrow{ c } \times \overrightarrow{ b })-\overrightarrow{ b }-\overrightarrow{ c })$ is equal to :
→
$\mathop {\lim }\limits_{x \to 0} \frac{{{x^2} - \tan 2x}}{{\tan x}} = $
→