 Let's take a look at an example of a problem in which we're going to use logarithmic differentiation to find the derivative of the expression. So remember with logarithmic differentiation we're going to use the laws of logarithms to first expand the given expression and then we employ implicit differentiation to actually find the derivative. So let's start by taking the natural log of both sides. There is a reason we use the natural log as opposed to the regular log in some other base. So we would have the natural log of y on the left. Now on the right when we take the natural log of that quotient, remember it's going to be the natural log of the numerator minus the natural log of the denominator. And that's according to the laws of logarithms. Now I'm going to rewrite that square root as an expression the x squared plus one raised to the one half. Now we can use the laws of logarithms again on the right side. Remember that each of those exponents can pop to the front. So now we're there and we can use implicit differentiation to now do the derivative. Now this is why we use natural log because the derivative of natural log is simply one over the expression times the derivative of the expression. It's a nice easy rule to use. So the derivative of natural log of y on the left would be one over y times dy dx. Remember with implicit differentiation you're taking the derivative with respect to x. Until any time you have a derivative of a y term, remember that we insert dy dx to denote that we are taking the derivative with respect to x. Now on the right side everything's in terms of x. So really it would be dx dx that we'd be multiplying by but that of course is one. So derivative of two times the natural log of the quantity x minus two, we keep the two as the coefficient and it's one over x minus two minus. Now we have the one half. Now for the derivative of natural log of x squared plus one that's going to require the chain rule so we would have one over x squared plus one but then we need to multiply by the inside derivative which simply would be two x. So we can start simplifying on that right side of the equal sign. You'll notice that the twos over here cancel out. So we can just simply condense this a little bit. Now for our final step remember that we're trying to solve for dy dx that denotes our derivative. So what we'll need to do is multiply both sides by y so that we have dy dx by itself on the left. So that gives us dy dx equals y times two over x minus two minus x over x squared plus one. It's perfectly acceptable to leave our answer like that. Remember that with implicit differentiation we often had both x's and y's in our answer. Another option and you will see this sometimes as well is to substitute in place of this y the expression you were given at the very beginning for y which in this case was x minus two that quantity squared over the square root of x squared plus one. Really either way is fine. Sometimes you'll see it one way, sometimes you'll see it another way. You could even, if you were to make the substitution, keep going and maybe distribute over the big parenthesis quantity and maybe simplify a little bit more but it's not really necessary. So it would be perfectly fine to stop at that first line where you see the answer as y times the quantity two over x minus two minus x over x squared plus one.