JEE Main 2023MathematicsDifferentiationMediumMCQ

JEE Main 2023 — Differentiation Question with Solution

JEE Main 2023 (01 Feb Shift 2)

Question

If yx=xx,x>0, then y"2-2y'2 is equal to :

Choose an option

Show full solutionCorrect option: C
Correct answer
C4loge22-2

Step-by-step explanation

Given, y=xx

⇒lny=xlnx

Differentiate both sides w.r.t. x

1y.y'=1+lnx

⇒y'=y1+lnx .......(1)

⇒y'=xx1+lnx

at x=2⇒y=4

So, y'(2)=41+ln2 .......(2)

Now differentiate the equation 1,

y"=y'1+lnx+yx

⇒y"(2)=y'1+ln2+2

⇒y"(2)-y'(2)=y'(ln2)+2

⇒y"(2)-2y'(2)=ln2-1y'(2)+2

=4(ln2-1)(ln2+1)+2

=4ln22-2

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About this question

This is a previous-year question from JEE Main 2023, 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.