Differentiation x^x

Differentiating y=x^x is simple; however it seems to have everyone puzzled. I have seen some top ranking bloggers educated in the finest university in the land with top grades make a mess of this one.

The best way to bring the power of x down is to use the log trick, by taking the log on both sides. As you can see on the RHS, the x then comes down and we have something we can work with.

Just use d/dx on both sides to keep the equation balanced. The RHS is the easy one as you can use the product rule to solve it. However, the LHS requires some subtle manipulation still.

We need to differentiate ln y with respect to y hence we need d/dy ln y. There is a simple way to change that by using the following mechanism dy/dx × d/dy = d/dx. As you can see, the dy terms cancels out so the expression remains d/dx in essence, however it gives us a way to differentiate with respect to y.

To differentiate the RHS I am using the product rule where f(x) = x, and g(x) = ln x, therefore the function is essentially y = f(x) × g(x).

Since g(x) = ln x, and its derivative is 1/x, this part is simple.

Using the product rule formula the RHS part works out like this and we end up with a simple expression of 1+ln x.

Replacing the RHS with what we previously calculated gives us the expression above. Obviously d/dy ln y = 1/y on the LHS.

The y moves to the RHS and we can substitute for it as well to give the result. Therefore the derivative of x^x is (x^x)(1 + ln x).

This Article Continues...