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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

If y=x^ n x+x( x)^ n , then (d y)/(d x) is equal to

If y=xnlog⁡x+x(log⁡x)ny=x^{n} \log x+x(\log x)^{n}, then dydx\frac{d y}{d x} is equal to

Options

  1. Axn−1(1+nlog⁡x)+(log⁡x)n−1[n+log⁡x]x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]
  2. Bxn−2(1+nlog⁡x)+(log⁡x)n−1[n+log⁡x]x^{n-2}(1+n \log x)+(\log x)^{n-1}[n+\log x]
  3. Cxn−1(1+nlog⁡x)+(log⁡x)n−1[n−log⁡x]x^{n-1}(1+n \log x)+(\log x)^{n-1}[n-\log x]
  4. DNone of the above

Correct answer

A. xn−1(1+nlog⁡x)+(log⁡x)n−1[n+log⁡x]x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]

Step-by-step solution

Given, y=xnlog⁡x+x(log⁡x)ny=x^{n} \log x+x(\log x)^{n}dydx=nxn−1log⁡x+xn⋅1x+xn(log⁡x)n−1(1x)+1⋅(\logx)n=xn−1(1+nlog⁡x)+(log⁡x)n−1[n+log⁡x]\begin{array}{c}\frac{d y}{d x}=n x^{n-1} \log x+x^{n} \cdot \frac{1}{x}+x n(\log x)^{n-1} \left(\frac{1}{x}\right)\\ +1 \cdot(\logx)^{n} \\ =x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x] \end{array}
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