Question

If $\mathrm{y}={\mathrm{x}}^{\mathrm{x}}$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is

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

Answer 1

The correct option is C

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

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

The given function is $\mathrm{y}={\mathrm{x}}^{\mathrm{x}}$

Taking $\log$ on both sides,

$\log y=x\log x$

Differentiating with respect to $\mathrm{x}.$

$\begin{array}{rcl}\frac{1}{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \log x+\mathrm{x}\times \frac{1}{\mathrm{x}}\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{log}\mathbf{}\mathbf{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{,}\mathbf{}\frac{\mathbf{d}\mathbf{(}\mathbf{a}\mathbf{·}\mathbf{b}\mathbf{)}}{\mathbf{dx}}\mathbf{=}\mathbf{a}\frac{\mathbf{db}}{\mathbf{dx}}\mathbf{+}\mathbf{b}\frac{\mathbf{da}}{\mathbf{dx}}\mathbf{]}\\ & =& 1+\mathrm{logx}\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& \mathrm{y}[1+\mathrm{logx}]\\ & =& {\mathrm{x}}^{\mathrm{x}}[1+\mathrm{logx}]\end{array}$

Hence, the correct option is C.