How do you differentiate x^x?

There are two ways we can find the derivative of x^x.

It's important to notice that this function is neither a power function of the form x^k nor an exponential function of the form b^x, so we can't use the differentiation formulas for either of these cases directly.

Method 1
1) Let y=x^x, and take logarithms of both sides of this equation:

ln(y)=ln(x^x)

2) Using properties of logarithmic functions, we can rewrite this as:

ln(y)=x.ln(x)

3) Then, differentiating both sides with respect to x and using the chain rule on the LHS and product rule on the RHS, this gives us:

1/y.dy/dx=ln(x)+1

4) Rearranging, we have:

dy/dx=y.(ln(x)+1). That is, dy/dx=x^x(ln(x)+1).

Method 2
1) Write x^x=e^(ln(x^x))=e^(x.ln(x)), using the properties of the exponential and logarithmic functions.

2) Now, d/dx(x.ln(x))=ln(x)+1 by the product rule. Hence, d/dx(e^(x.ln(x)))=(ln(x)+1).(e^(x.ln(x)) by the chain rule, and using the fact that the derivative of e^[f(x)]=f'(x).e^[f(x)] for any differentiable function f(x).

3) Finally, rewriting e^(x.ln(x)) as x^x gives d/dx(x^x)=x^x.(ln(x)+1), as with the first method.