JEE Main 2018MathematicsDifferentiationEasyMCQ

JEE Main 2018 — Differentiation Question with Solution

JEE Main 2018 (15 Apr)

Question

If x2+y2+siny=4, then the value of d2ydx2 at the point -2, 0 is :

Choose an option

Show full solutionCorrect option: A
Correct answer
A-34

Step-by-step explanation

Given x2+y2+siny=4

Differentiating both sides with respect to x, we get

2x+2y+cosydydx=0

⇒dydx=-2x2y+cos⁡y

At -2, 0, dydx= 41=4

Also, 2y+cosydydx+2x=0

Again differentiating with respect to x, we get

2y+cosyd2ydx2+2-sinydydx2+2=0

At -2, 0, d2ydx2+2-042+2=0

⇒d2ydx2=-34

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About this question

This is a previous-year question from JEE Main 2018, 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.