Differentiate the following with respect to x : $\left( {{x}^{x}} \right)$.

Hint:We cannot use the general formula of differentiation to solve this question. So, we will consider $y={{x}^{x}}$ and take $\ln $ on both sides and then differentiate y with respect to x to obtain the required answer. We will be using the chain rule, $\dfrac{d}{dx}f\left( g\left( x \right) \right)=\dfrac{d}{dx}f\left( g\left( x \right) \right).\dfrac{d}{dx}g\left( x \right)$ as well as the product rule given by $\dfrac{d}{dx}\left( u.v \right)=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx}$ to find the derivative.

Complete step by step answer:
In this question, we can see that $\left( {{x}^{x}} \right)$ is not in the form of a power function, ${{x}^{k}}$ and not in the exponential function form of ${{e}^{x}}$, so the formula of differentiation cannot be used here and we need to solve the given function using a different method.
So, let us consider $y={{x}^{x}}$ and assume that, $x>0$. We will now take the logarithm function on both the sides. So, we will get as,
$\ln y=\ln {{x}^{x}}$
Now, we know that $\log {{a}^{b}}=b\log a$. We will apply this property in the above equality and get,
$\ln y=x\ln x$
We will now differentiate both sides with respect to x. We will use chain rule, given by $\dfrac{d}{dx}f\left( g\left( x \right) \right)=\dfrac{d}{dx}f\left( g\left( x \right) \right).\dfrac{d}{dx}g\left( x \right)$ on the left side and product rule, given by $\dfrac{d}{dx}\left( u.v \right)=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx}$ on the right side. So, we get,
$\begin{align}
  & \dfrac{1}{y}\times \dfrac{dy}{dx}=\ln x+x\times \dfrac{1}{x} \\
 & \Rightarrow \dfrac{1}{y}\times \dfrac{dy}{dx}=\ln x+1 \\
\end{align}$
Taking y to the right hand side, we get,
$\dfrac{dy}{dx}=y\left( \ln x+1 \right)$
We will now substitute the value of y as, $y={{x}^{x}}$. So, we get the above equality as,
$\dfrac{dy}{dx}={{x}^{x}}\left( \ln x+1 \right)$
Therefore, we get the answer as ${{x}^{x}}\left( \ln x+1 \right)$.

Note:
 We can also solve this question by using an alternate method. We know that $\log {{a}^{b}}=b\log a\text{ and }{{e}^{\log x}}=x$ . Now, using these, we can use the exponential rule, ${{e}^{b\ln a}}={{a}^{b}}$ to express ${{x}^{x}}\Rightarrow {{e}^{x\ln x}}$ . Now, we can apply chain rule as ${{e}^{x\ln x}}.\dfrac{d}{dx}\left( x\ln x \right)$ and then proceed with product rule as usual. Both ways, we will get the same result.