Q. If $y = x^{n}\log{x + x\left( \log x \right)^{n},\ then\frac{dy}{dx}\ is\ equal\ to\ }$
A) $x^{n - 1}\left( 1 + n\ \log\ x \right) + {(\log{x)}}^{n - 1}\lbrack n + \log{x\rbrack}$
B) $x^{n - 2}(1 + n\log{x) + {(\log{x)}}^{n - 1}\lbrack n + \log{x\rbrack}}$
C) $x^{n - 1}(1 + n\log{x) + {(\log x)}^{n - 1}\lbrack n - \log{x\rbrack}}$
D) None of the above
Explanation
(a)
Given, $y = x^{n}\log{x + x{(\log{x)}}^{n}}$
On differentiating w.r.t. $x,\$we get
$\frac{dy}{dx} = nx^{n - 1}\log{x + x^{n}.\frac{1}{x} + xn{(\log{x)}}^{n - 1}\left( \frac{1}{x} \right) + 1.{(\log{x)}}^{n}}$
$= x^{n - 1}(1 + n\log{x) + {(\log{x)}}^{n - 1}\lbrack n + \log{x\rbrack}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.