Question

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

Answer 1

Hint: To solve this problem we need to divide the given expression in two parts i.e. first \[{{\left( \log x \right)}^{x}}\] and second is ${{x}^{\log x}}$. Now we will assume both the parts to be equal to some variable, say y then apply log on both the sides to solve it, after that we will differentiate both the parts separately and then we will combine the result to get the final result. We will assume both the parts to be equal to some variable, say y then apply log on both the sides to solve it. We will also be applying the chain rule and product rule to solve this question, the chain rule says, the derivative of
f(g(x)) = f'(g(x)).g'(x). and the product rule says: (fg)’ = f g’ + f’ g.

Complete step by step answer:
We have to differentiate,
${{\left( \log x \right)}^{x}}+{{x}^{\log x}}$ with respect to x,
First, we will divide the above expression in two parts first is \[{{\left( \log x \right)}^{x}}\] and second is ${{x}^{\log x}}$ and now we will suppose \[{{y}_{1}}={{\left( \log x \right)}^{x}}\] and ${{y}_{2}}={{x}^{\log x}}$,
We will now differentiate ${{y}_{1}}$ first with respect to $x$, as
\[{{y}_{1}}={{\left( \log x \right)}^{x}}\]
Taking log on both sides, we get
$\log \left( {{y}_{1}} \right)=\log {{\left( \log x \right)}^{x}}$
And we know that $\log {{x}^{n}}=n\log x$,
$\log \left( {{y}_{1}} \right)=x\log \left( \log x \right)$
Differentiating both sides with respect to x, we get
\[\dfrac{d\log \left( {{y}_{1}} \right)}{dx}=\dfrac{d\left( x\log \left( \log x \right) \right)}{dx}\]
Now we will apply chain rule as well as product rule on the RHS of the above equation,
According to the chain rule the derivative of
f(g(x)) = f'(g(x)).g'(x).
and according to the product rule, (fg)’ = f g’ + f’ g,
so using the above properties, we get
\[\dfrac{1}{{{y}_{1}}}\dfrac{d{{y}_{1}}}{dx}=x.\dfrac{1}{\log x}.\dfrac{1}{x}+\log \left( \log x \right)\]
\[\begin{align}
  & \dfrac{d{{y}_{1}}}{dx}={{y}_{1}}\left( \dfrac{1}{\log x}+\log \left( \log x \right) \right) \\
 & \dfrac{d{{y}_{1}}}{dx}={{\left( \log x \right)}^{x}}\left( \dfrac{1}{\log x}+\log \left( \log x \right) \right) \\
\end{align}\]

Now we will differentiate ${{y}_{2}}$ first with respect to $x$, as
${{y}_{2}}={{x}^{\log x}}$
Taking log on both sides, we get
$\log \left( {{y}_{2}} \right)=\log \left( {{x}^{\log x}} \right)$
And we know that $\log {{x}^{n}}=n\log x$,
$\begin{align}
  & \log \left( {{y}_{2}} \right)=\log \left( {{x}^{\log x}} \right) \\
 & \log \left( {{y}_{2}} \right)=\log x.\log \left( x \right) \\
 & \log \left( {{y}_{2}} \right)={{\left( \log x \right)}^{2}} \\
\end{align}$
Differentiating both sides with respect to x and applying chain rule on RHS, we get
\[\begin{align}
  & \dfrac{d\left( \log {{y}_{2}} \right)}{dx}=\dfrac{d\left( {{\left( \log x \right)}^{2}} \right)}{dx} \\
 & \dfrac{1}{{{y}_{2}}}\dfrac{d{{y}_{2}}}{dx}=2\log x.\dfrac{1}{x} \\
\end{align}\]
\[\begin{align}
  & \dfrac{d{{y}_{2}}}{dx}={{y}_{2}}\times \left( 2\log x.\dfrac{1}{x} \right) \\
 & \dfrac{d{{y}_{2}}}{dx}={{x}^{\log x}}\times \left( 2\log x.\dfrac{1}{x} \right) \\
\end{align}\]
Now we know that,
$\dfrac{d\left( {{\left( \log x \right)}^{x}}+{{x}^{\log x}} \right)}{dx}=\dfrac{d{{y}_{1}}}{dx}+\dfrac{d{{y}_{2}}}{dx}$
Hence we get,
$\begin{align}
  & \dfrac{d\left( {{\left( \log x \right)}^{x}}+{{x}^{\log x}} \right)}{dx}={{\left( \log x \right)}^{x}}\left( x.\dfrac{1}{\log x}+\log \left( \log x \right) \right)+{{x}^{\log x}}.\left( 2\log x.\dfrac{1}{x} \right) \\
 & \\
\end{align}$

Above expression is our required answer.

Note: While solving questions for differentiation which involves \[{{\left( {{f}_{1}}\left( x \right) \right)}^{{{f}_{2}}\left( x \right)}}\] (where ${{f}_{1\,}}\,and\,{{f}_{2}}$ are any two functions of $x$ always remember that these type of questions can be easily solved by taking log on both sides of the equation. And also read and learn basic properties of logarithmic functions in order to solve problems involving them.