Question

If $y=x^n\log x+x{\left(\log x\right)}^n$, then $\frac{dy}{dx}$ is equal to

• A. $x^{n-1}(1+n\log x)+{\left(\log x\right)}^{n-1}[n+\log x]$
• B. $x^{n-2}(1+n\log x)+{\left(\log x\right)}^{n-1}[n+\log x]$
• C. $x^{n-1}(1+n\log x)+{\left(\log x\right)}^{n-1}[n-\log x]$
• D. None of the above

Answer 1

The correct option is A $x^{n-1}(1+n\log x)+{\left(\log x\right)}^{n-1}[n+\log x]$

Given, $y=x^n\log x+x{\left(\log x\right)}^n$
$\frac{dy}{dx}=n{x}^{n-1}\log x+x^n\cdot \frac{1}{x}+xn{\left(\log x\right)}^{n-1}\left(\frac{1}{x}\right)+1\cdot {\left(\log x\right)}^n$
$=x^{n-1}(1+n\log x)+{\left(\log x\right)}^{n-1}[n+\log x]$