Skip to main content
Back to DIFFERENTIATION

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy
1 of 1
Analyzing +4 | -1
The function f(x)f(x) given by f(x)= arc  sin(2x1+x2)f(x)=\,arc\,\,sin \left(\frac{2 x}{1+x^2}\right), is
View answer and explanation
Correct option: B
such that f′(x)={21+x2,    −1<x<1−21+x2,    ∣x∣>1f'(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{2}{{1 + {x^2}}},\,\,\,\, - 1 < x < 1}\\ {\frac{{ - 2}}{{1 + {x^2}}},\,\,\,\,|x| > 1} \end{array}} \right.
Option B is correct because the derivative of the arcsin function needs to be evaluated piecewise due to the range restriction of the arcsin function, resulting in different derivatives for |x| < 1 and |x| > 1.

More Questions from DIFFERENTIATION

Practice All Chapter MCQs →
If f(x)=∣x∣∣tan⁡x∣f(x)=|x|^{|\tan x|}, then f′(−π6)f^{\prime}\left(\frac{-\pi}{6}\right) is equal to Let xx and yy be two real numbers such that x>0x>0 and xy=1x y=1. The minimum value of x+yx+y is Let f(x)=[x]sin⁡(πx)f(x)=[x] \sin (\pi x) where [x][x] denotes the integer part of xx. Then at x=k,kx=k, k being an integer, the left derivative of f(x)f(x) is If y=sec⁡−1[cosecx]+cosec−1[sec⁡x]+sin⁡−1[cos⁡x]+cos⁡−1[sin⁡x],y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x], then dydx\frac{{dy}}{{dx}} is equal to ddxsin⁡−1(3x−4x3)\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right) is equal to If z=y+f(v),z = y + f\left( v \right), where v=(xy),v = \left( {\frac{x}{y}} \right), then v∂z∂x+∂z∂yv\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} is
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →