Question

If $y=\log x^x$ , then the value of $\frac{dy}{dx}$ is

• A. $x^x\left(1+\log x\right)$
• B. $\log ex$
• C. $\log \frac{e}{x}$
• D. $\log \frac{x}{e}$

Answer 1

The correct option is B

$\log ex$

Find the value of $\frac{dy}{dx}$:

Given,

$$\begin{gathered} y=\log x^x \\ y=x\log x\left[\because log{a}^b=bloga\right] \end{gathered}$$

Now differentiate with respect to $x$.

Then,

$\begin{array}{rcl}\frac{dy}{dx}& =& \log x+x\left(\frac{1}{x}\right)\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{u}\mathbf{v}\mathbf{}\mathbf{=}\mathbf{}\mathbf{u}\mathbf{\text{'}}\mathbf{v}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{u}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{]}\\ & =& \log x+1\\ & =& \log x+\log e\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{log}\mathbf{}\mathbf{e}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{]}\\ & =& \log ex\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{log}\mathbf{}\mathbf{a}\mathbf{b}\mathbf{}\mathbf{=}\mathbf{}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{b}\mathbf{]}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Hence, the correct option is B.