Question

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

Answer 1

Answer: (A)

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