QUESTION

# The value of $\dfrac{d}{{dx}}\left( {{x^x}} \right)$ is equal to:A. ${x^x}\log \left( {\dfrac{e}{x}} \right)$B. ${x^x}\log ex$C. ${x^x}\left( {1 + \log x} \right)$D. ${x^x}\log x$

Hint: First of all, apply logarithm to the function to obtain a simple equation. Then use the product rule of derivatives to find the derivative of the given function. So, use this concept to reach the solution of the given problem.

Let $y = {x^x}$
$\Rightarrow \log y = \log {x^x}$
We know that $\log {a^b} = b\log a$
$\Rightarrow \log y = x\log x$
Differentiating on both sides w.r.t $x$, we get
$\Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) = \dfrac{d}{{dx}}\left( {x\log x} \right)$
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x\dfrac{d}{{dx}}\left( {\log x} \right) + \log x\dfrac{d}{{dx}}\left( x \right) \\ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} + \log x \times 1 \\ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = 1 + \log x \\ \Rightarrow \dfrac{{dy}}{{dx}} = y\left( {1 + \log x} \right) \\ \therefore \dfrac{{dy}}{{dx}} = {x^x}\left( {1 + \log x} \right){\text{ }}\left[ {\because y = {x^x}} \right] \\$
Therefore, the derivative of ${x^x}$ is ${x^x}\left( {1 + \log x} \right)$.
Thus, the correct option is C. ${x^x}\left( {1 + \log x} \right)$.
Note: The product rule states that if $f\left( x \right)$ and $g\left( x \right)$ are both differentiable, then $\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]$. Remember the derivative of ${x^x}$ as a formula which will be useful to solve higher derivative problems.