If $y = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x + {\sec ^{ - 1}} + {\rm{cose}}{{\rm{c}}^{ - 1}}x,$ then $\frac{{dy}}{{dx}}$ is equal to

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

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Understanding +4 | -1

y = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x + {\sec ^{ - 1}} + \mathrm{cose}{\mathrm{c}^{ - 1}}x,

then

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

is equal to

View answer and explanation

Correct option: C

Option C (0) is correct because the expression simplifies to a constant, and the derivative of a constant is zero.

Practice All Chapter MCQs →

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