Question

How do you find the derivative of \[f\left( x \right)={{x}^{\log x}}\]?

Answer 1

Hint: From the given question, we have been asked to find the derivative of \[f\left( x \right)={{x}^{\log x}}\]. We can find the derivative of the given question by using some basic formula of differentiation. From the above given question, we have been asked to find the derivative of \[f\left( x \right)={{x}^{\log x}}\]

Complete step-by-step solution:
First of all, let us assume the given function as some variable \[y\].
By assuming the given function with variable \[y\], we get \[y={{x}^{\log x}}\]
Now, apply logarithm on both sides of the equation to get the equation more simplified.
By applying the logarithm on the both sides of the equation, we get the equation as
\[\log y=\log x\log x\]
\[\Rightarrow \log y={{\left( \log x \right)}^{2}}\].
Now, apply differentiation on both sides of the above equation.
By applying differentiation on both sides of the equation, we get \[\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{2}{x}\log x\]
Now, shift \[\dfrac{1}{y}\] from the left hand side of the equation to the right hand side of the equation.
By shifting \[\dfrac{1}{y}\] from left hand side of the equation to the right hand side of the equation, we get \[\dfrac{dy}{dx}=\dfrac{2y}{x}\log x\]
We already assumed that \[y={{x}^{\log x}}\] at the starting of the problem.
Now, substitute \[y={{x}^{\log x}}\] in the above equation to get the final derivative of the given question.
By substituting, we get
\[\dfrac{dy}{dx}=\dfrac{2{{x}^{\log x}}}{x}\log x\]
\[\Rightarrow \dfrac{dy}{dx}=2{{x}^{\log x-1}}\log x\]
Hence, a derivative for the given question is found.

Note: We should be well aware of the formulae of differentiation. Also, we should know the usage of the formulae of differentiation. Assumptions and substitutions must be made very carefully by us to get the correct derivative for the given question. Calculation must be done very carefully. The derivatives can be easily done by remembering some formulae like $\dfrac{d}{dx}\log x=\dfrac{1}{x}$, $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ and many more.