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Web AppCalculating the derivative of x^x is a very simple task, but it may be hard to find the general idea on your own, so here it is. We will need the following formula:
a^b = \l(e^{\ln(a)}\r)^b = e^{b\ln(a)}(where “\ln” denotes the natural lnarithm, which is often denoted “\ln” in non-mathematical literature). The trick is to rewrite x^x using the formula above (by setting a = b = x) and then differentiate the rest according to the normal rules of differentiation:
\l(x^x\r)’ = \l(e^{x\ln(x)}\r)’ = e^{x\ln(x)}\l(x\ln(x)\r)’\,,where we used (e^{f(x)})’ = e^{f(x)}f’(x). Now we will rewrite e^{x\ln(x)} back to x^x (it is the same thing we started with) and differentiate the second part using (u(x)v(x))’ = u’(x)v(x) + u(x)v’(x):
e^{x\ln(x)}\l(x\ln(x)\r)’ = x^x\l(1⋅\ln(x) + x⋅\frac{1}{x}\r) = x^x\l(\ln(x)+1\r)\,.And there you have it: (x^x)’ = x^x\l(\ln(x)+1\r).
