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# How do you differentiate x ^ x ?

Derivative in mathematics is the rate of change of a function with respect to a variable.

Derivatives are very important to the solution of problems in calculus.  To differentiate a function one must know the rules of derivatives.

The derivative of a constant function is always zero.

Let $y\,=\,f\,(x)$  is any function.

We can explain it as the measure of the rate at which the value of  changes with respect to the change of the variable $x$. It is read as the derivative of function $f$  with respect to the variable $x$. Now the question is how we can differentiate  $\,{{x}^{x}}?$

So, regarding this question we can say that there are two ways to find the derivative of ${{x}^{x}}$ .

There is an important point to be noticed here that this function is neither a power function of the form $x{{\,}^{\hat{\ }}}\,k$ where $k$ is any constant nor an exponential function of the form $b\,\hat{\ }x$  where b is a constant, so we can’t use the differentiation formulas for either of these cases directly.

Symbol of derivative

The symbol for denoting the derivative of a function is  $f'(x)$  or we can also denote it by $\frac{{d(y)}}{{dx}}$

Differentiating $y\,=\,x\,\hat{\ }\,x$ is simple but tricky.

## Method 1

The best way to solve it is to bring the power down and we can bring the power down only by taking logarithm (ln) to the natural base e on both sides.

$Let\,y\,=\,x\,\hat{\ }\,x$

1Taking log both sides ,we get0

$ln\,y\,=\,ln\,x$

Now taking derivative on both sides, we get

$\frac{{d(\ln \,y)}}{{dx}}\,=\,\frac{{d\,(x\,\ln \,x)}}{{dx}}$

To differentiate the right hand side, we can use the product rule

Product rule says

• First function $d/dx$ second function   + second function $d/dx$ first function
• Also, According to properties of derivatives: derivative of $\log \,x$ is  $1/x$  and derivative of $x$ is 1.

So, applying this rule above we get

$\begin{array}{l}1/y\,\,\frac{{dy}}{{dx}}\,\,=\,x\,\frac{d}{{dx}}\,(ln\,x)\,+\,ln\,x\,\frac{d}{{dx}}\,(x)\\1/y\,\,\frac{{dy}}{{dx}}\,\,=\,\,x(1/x)\,\,+\,\,ln\,x\,(1)\\1/y\,\,\frac{{dy}}{2}\,\,=\,\,1+\,ln\,x\\\frac{{dy}}{{dx}}\,\,=\,y\,(1\,+\,ln\,x)\\\frac{{dy}}{{dx}}\,\,=\,\,x\hat{\ }x\,(1+ln\,x)\,\,\,[as\,the\,value\,of\,y\,=\,x\,\hat{\ }\,x]\end{array}$

So , derivative of  ${{x}^{x}}\,is\,{{x}^{x}}(1\,+\,ln\,x)$ by applying logarithm method as this can’t be differentiated with the usual differential methods of derivatives such as product rule,  chain rule, quotient rule etc.

## Method 2

There is one more method of finding its derivative by using exponential and logarithm method.

We know that  ${{e}^{{ln\,x\,}}}\,=\,x$

So, this is same as  ${{x}^{x}}\,as\,\,({{e}^{{ln\,x\,}}})\,\hat{\ }x$

Which is also same as  $e\,\hat{\ }\,(x\,ln\,(x))\,or\,\,{{e}^{{x\,ln\,x\,}}}$ [using the properties of exponential and logarithmic functions]

We know that,

Hence , by using the formula $\frac{d}{{dx}}\,{{e}^{u}}\,=\,{{e}^{u}}\,\frac{{du}}{{dx}}$

$\frac{d}{{dx}}(\,{{e}^{{x\,lnx}}}\,)\,=\,{{e}^{{xlnx}}}\,\,\frac{d}{{dx}}\,(xln\,x)$

$\frac{d}{{dx}}(\,{{e}^{{x\,lnx}}}\,)\,\,=\,{{e}^{{xlnx}}}\,\,(ln\,(x)\,+\,1)$ [using product rule and also derived in method 1]

Finally rewriting it will give the same answer i.e.

$d/\,dx\,\,(\,{{x}^{x}}\,)\,\,=\,(1\,+\,ln\,x\,)\,\,[as\,{{e}^{{x\,ln\,x\,}}}\,is\,nothing\,buy\,\,\,{{x}^{x}}]$