Question

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

• A. ${\mathrm{x}}^{\mathrm{n}-1}(1+\mathrm{nlog}\mathrm{x})+(\log \mathrm{x}{)}^{\mathrm{n}-1}(\mathrm{n}+\log \mathrm{x})$
• B. ${\mathrm{x}}^{\mathrm{n}-2}(1+\mathrm{nlog}\mathrm{x})+(\log \mathrm{x}{)}^{\mathrm{n}-1}(\mathrm{n}+\log \mathrm{x})$
• C. ${\mathrm{x}}^{\mathrm{n}-1}(1+\mathrm{nlog}\mathrm{x})+(\log \mathrm{x}{)}^{\mathrm{n}-1}(\mathrm{n}-\log \mathrm{x})$
• D. None of these

Answer 1

The correct option is A

${\mathrm{x}}^{\mathrm{n}-1}(1+\mathrm{nlog}\mathrm{x})+(\log \mathrm{x}{)}^{\mathrm{n}-1}(\mathrm{n}+\log \mathrm{x})$

Finding the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$:

The given differential function is $\mathrm{y}={\mathrm{x}}^{\mathrm{n}}\log \mathrm{x}+\mathrm{x}{(\log \mathrm{x})}^{\mathrm{n}}$.

Differentiate the given equation with respect to $x$.

$\begin{array}{c}\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{nx}}^{\mathrm{n}-1}\mathrm{logx}+{\mathrm{x}}^{\mathrm{n}}\left(\frac{1}{\mathrm{x}}\right)+{\left(\mathrm{logx}\right)}^{\mathrm{n}}+\mathrm{xn}{\left(\mathrm{logx}\right)}^{\mathrm{n}-1}\left(\frac{1}{\mathrm{x}}\right)\left[\because \frac{d\left(\mathrm{xy}\right)}{\mathrm{dx}}=x\frac{\mathrm{dy}}{\mathrm{dx}}+y\frac{d\left(x\right)}{\mathrm{dx}},\frac{d\left(x^n\right)}{\mathrm{dx}}={\mathrm{nx}}^{n-1},\frac{\mathrm{dlogx}}{\mathrm{dx}}=\frac{1}{x}\right]\\ ={\mathrm{x}}^{\mathrm{n}-1}\left(\mathrm{nlogx}+1\right)+{\left(\mathrm{logx}\right)}^{\mathrm{n}-1}\left(\mathrm{logx}+\mathrm{n}\right)\end{array}$

Hence, the correct option is A.