 In this video we will provide the solution to question number 14 for practice exam number 3 for Math 12-10. We're asked to use logarithmic differentiation to find the derivative of the function y equals x to the x power. So when it comes to logarithmic differentiation, we take our function, our equation y equals x to the x. We have to take the natural log of both sides. Take the natural log of y on the left-hand side and take the natural log of x to the x on the right-hand side. The whole point of logarithmic differentiation is to expand these expressions using logarithmic properties. So the natural log of x to the x, because of the laws of logarithms, we can bring that x out as a coefficient. So we get x times the natural log of x and now we're going to take the derivative of both sides. So we take the derivative of the natural log of y and we take the derivative of x times the natural log of x. And we're taking the derivative with respect to x. So I mean, be aware we're taking the derivative with respect to x in both of these situations. That will be clear on the right-hand side, but it's important you know this on the left-hand side because then when you take the derivative on the left-hand side, you're going to get y prime over y. The nice thing about logarithmic differentiation, if you have an explicit function relationship like you do here, then the left-hand side when you take the derivative will always look like y prime over y. The right-hand side will depend on the function itself. The product rule is going to come into play here. You have to take the derivative of x times that by the natural log and then we're going to take the derivative of the natural log of x times that by x. So we see that y prime over y, this is going to equal, well the derivative of x is just 1, so we get 1 times the natural log of x. And then the derivative of the natural log of x is going to be 1 over x. Like so, simplifying the right-hand side, we get the natural log of x plus 1. This is equal to y prime over y. Many of us stop at this moment thinking that we're done, but no, we didn't find the derivative yet. We found y prime divided by y. So to solve for the derivative, we need to times both sides by y so that they cancel on the left-hand side and we just have a y prime. But then we get that y prime equals y times the natural log of x plus 1. As the original function was given explicitly, what we need to do is write the derivative explicitly as well. So we didn't make the substitution in right here. And so we see that the derivative y prime is going to equal x to the x times the natural log of x plus 1, which is then the correct derivative of our function which we found logarithmically.