JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: A

1-\frac{1}{x^{2}}

The correct answer is A (

1-\frac{1}{x^{2}}

) because it is the result of differentiating the given expression after simplifying it.

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

x

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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)

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