 When we combine the derivative of a log with the rules of logs, we find a new way of finding the derivatives of very complicated functions. So as a reminder, we have the following four rules of logs. The log of a product is the sum of the individual logs. The log of a quotient, a over b, is log a minus log b. The log of a power, a to the n, is n log a, and the log of an nth root is 1 over n times log of a. So I suppose I want to find the derivative of this horrible expression. Remember, the type of function is determined by the last operation performed, and this is a root function. And that's important because logs can simplify roots. So let's hit both sides with a log, and since this is a log of a root, we can simplify it, bringing the index out front as a fraction one-third. But wait, there's more. Again, the type of function is determined by the last operation performed, and here we're taking the log of a product, and we can simplify that. But wait, there's still more. The type of operation is determined by the last operation performed, and here we have the log of a sum and the log of a difference, and we check our rules of logs, and we see that the log of a sum isn't there, and that means we can't simplify this expression any further. Now we can use implicit differentiation. On the left hand side, we have log something, and the derivative of log something is 1 over something times the derivative of something. On the right hand side, we have constant multiple of a function, and the derivative of constant multiple of a function is going to be one-third times the derivative. And again, we take a look at our function, and we see that it's a sum, and so the derivative of a sum is the sum of the derivatives, and we can differentiate each part separately. So we'll differentiate log of x squared plus 5, which is the log of something, and so our derivative will be 1 over something times our derivative of the something, and the derivative of log 2x minus 7, that's the derivative of log something, is going to be 2 over 2x minus 7. Since we want to find the derivative, we'll multiply both sides by y, and finally, since our function was given entirely in terms of x, we'll want to make sure our derivative is also given entirely in terms of x, so we'll replace y with what it's equal to. How about 5 to power x? It's vitally important to notice that this is not x raised to some constant power, for which we already have a derivative formula, this is some constant raised to a variable power, and we should expect that the derivative should look nothing like the derivative of x to the fifth, because our exponential function is nothing like x to the fifth. Well, this is a power, so let's hit both sides with the log, and we're taking the log of a power, and so we can simplify this. Now we'll differentiate both sides. On the left, we have the derivative of log something, and the derivative of log something will be 1 over something times derivative of something. On the right hand side, we have x log 5, and the thing to remember is log 5 is just a constant multiplier, so we can remove that to the front of the differentiation, and a derivative will be we want to find the derivative of f of x, and since the original function was given to us in terms of x only, we should make sure that the final answer is given in terms of x only, so we'll replace f of x with what it's equal to. The preceding approach leads to a completely useless formula for another derivative. The formula is so useless that I won't even put it up, because it's completely unusable on something like this, and yet our approach will still work, so we'll hit both sides with the log, then use our rules of logs to simplify, then I'll use implicit differentiation. On the left hand side, we have log something, so the derivative will be, and on the right hand side, we have the derivative of a product, which means we need to use the product rule. So we need the derivative of log sine of x, and so this function is a log something, so our derivative will be 1 over sine x cosine x. We also need the derivative of cosine x, and finally we want to find dy dx, so we'll multiply both sides by y, and get our final derivative expression.