Question

# Differentiate the following function with respect to x ${x^x}$

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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.