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Differentiate the following function with respect to x

Last updated date: 13th Jun 2024
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Hint: Differentiation of function means to compute the derivative of that function. A derivative is the rate at which output changes with respect to an input. We will suppose the given value as $\left( {y = {x^x}} \right)$.

Complete step by step solution:
$ y = {x^x} $
Taking log both sides, we will get
$ \log y = \log {x^x} $
We know that by the property of logarithm that $ {\text{log(m}}{{\text{)}}^{\text{a}}}{\text{ = a log m}} $
\[ \Rightarrow \log \,y = x{\text{ }}log{\text{ }}x\]
Differentiating this value with respect to x, we have
  \dfrac{1}{y}\dfrac{d}{{dx}}y = x\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}x \\
  \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} \times \dfrac{d}{{dx}}x + \log x \times 1 \\
  \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} \times 1 + \log x \\
  \dfrac{1}{y}\dfrac{{dy}}{{dx}} = 1 + \log x \\
\[\dfrac{1}{y}\,\,\dfrac{{dy}}{{dx}} = 1 + \log x\,\,\]
$ \dfrac{{dy}}{{dx}} = y\left( {1 + \log x} \right) $
\[\dfrac{{dy}}{{dx}} = {x^x}\left( {1 + \log \,x} \right)\] \[\left( {\because y = {x^x}} \right)\]

Note: In these types of questions students must take care while calculating the differentiation of \[log\,x\]. Usually students forget to calculate the derivative $ \left( {\dfrac{d}{{dx}}} \right) $ the value \[log\,x\] with respect to their function.