JEE Main 2019MathematicsDifferentiationEasyMCQ

JEE Main 2019 — Differentiation Question with Solution

JEE Main 2019 (12 Apr Shift 2)

Question

The derivative of tan-1sinx-cosxsinx+cosx with respect to x2, where x∈0,π2, is

Choose an option

Show full solutionCorrect option: A
Correct answer
A2

Step-by-step explanation

Given function can be written as

y=tan-1sinx-cosxsinx+cosx

⇒ y=tan-1tanx-1tanx+1

⇒ y=tan-1tanx-π4 given x∈0,π2
⇒ y=x-π4
Now dydx2=2dydx=2

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About this question

This is a previous-year question from JEE Main 2019, covering the Differentiation chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.