JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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Analyzing +4 | -1

The derivative of

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

with respect to

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

x=\frac{1}{2}

, is

View answer and explanation

Correct option: A

\frac{\sqrt{3}}{10}

Option A (

\frac{\sqrt{3}}{10}

) is correct because by substituting

x = tan(\theta)

and

x = sin(\phi)

, we can simplify both inverse trigonometric functions and then find the derivative at x=1/2.

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f(x)=|x|^{|\tan x|}

f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

x>0

x y=1

x+y

f(x)=[x] \sin (\pi x)

[x]

x

x=k, k

f(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],

\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}