Differentiate x^x

This can't be differentiated with the usual methods (chain rule, product rule). First we set y = xx, and our objective is to calculate dy/dx in terms of x.
To turn this function into more familiar functions, we log both sides:
ln y = ln(xx) = x ln x
where in the second equality we have used the logarithm rule ln(ab) = b ln a.
Then we differentiate both sides with respect to x. For the left hand side, we use the chain rule, dz/dx = (dz/dy) * (dy/dx), with z = ln y:
d(ln y)/dx = d(ln y)/dy * dy/dx = (1/y) * dy/dx
On the right hand side, we use the product rule, d(uv)/dx = u * dv/dx + v*du/dx, with u = x and v = ln x:
d(x ln x)/dx = x * (1/x) + ln x * 1 = 1 + ln x
So then we have
(1/y) * dy/dx = 1 + ln x
and our goal is to find dy/dx in terms of x only. To do this, we multiply both sides by y and substitute y = xx, leaving
dy/dx = xx (1 + ln x)

JT
Answered by Joel T. Maths tutor

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