 Now that we have some property to the derivative, let's go ahead and find some derivatives. So let's find the derivative of 1 over x squared using the definition. So again, we'll pull in our definition of the derivative, and we need to find our function at x plus h and at x. We'll substitute these values into our definition, and while we can try to evaluate this expression at h equal to 0, we should come to expect by now that we will get an indeterminate form and have to do some algebra. So we have a compound fraction. So let's multiply numerator and denominator by the denominators x squared times x plus h quantity squared. And we'll do a little algebra. Well, okay, we'll do a lot of algebra. But after all the dust settles, we'll get a derivative of minus 2 over x to the third power. Let's pause a moment and take stock of what we've found. So we just found that if f of x is equal to 1 over x squared, the derivative is minus 2 over x to the third power. In a previous problem, we found the derivative of x squared was equal to 2x. And we also found that the derivative of the square root of x was 1 over 2 square root of x. Well, let's take a look at these and see if there's any useful pattern that emerges. Now in order to analyze these, it's helpful if they are somewhat consistent. And so what we might start off with is writing each of these functions in exponential notation. So x to the second is x to the second. The square root of x and 1 over x squared can be rewritten in exponential form as well. But if we wrote the functions in exponential form, we should write the derivatives in exponential form as well. And if we look carefully, these three examples suggest an important derivative rule. If f of x equals x to the n, then f prime of x is nx to power n minus 1. In other words, our exponent comes down front as a coefficient and our power drops by 1. Now if we combine this with our other derivative rules, then what we get is a method of finding the derivative of any polynomial as well as a number of other types of functions. For example, suppose we want to find the derivative of x cubed plus 8x plus square root of x. Now, if the only tool you have is a hammer, everything looks like a nail. In this particular case, the only tool we have is the derivative of x to the n. So we have to make sure that everything looks like a nail. In this particular case, we want to make sure that as many things as possible look like something to the power n. So x cubed is already x to the n. 8x can be rewritten as 8x to the first and square root of x can be rewritten as x to power 1 half. Next, it's always useful to remember the type of function we're dealing with is determined by the last operation performed. In this case, the last thing we do when we try to evaluate x to the third plus 8x to the first plus x to power 1 half is we add. And so this function is a sum. And if we want to find the derivative of a sum, it'll be the sum of the derivatives. Now we know the derivative of x to the third and x to the power 1 half. But what about this derivative of 8 times x to the first? Well again, the last thing we do here is we multiply by 8, which is a constant, and so this is the derivative of a constant times a function. So what do we do with that? We have this result that the derivative of a constant times a function is the constant times the derivative of the function. So we can move that multiplier of 8 outside of the differentiation. And finally we can use our results. The derivative of x to the power 3 is, by our theorem, 3x to the power 2 minus 1. And the derivative of x to the 1 half will be 1 half x to the power minus 1 half. The derivative of x to the 1 will be 1x to the 0. And then we'll do a little bit of algebra and some cleanup. We'll give us our derivative. How about this derivative? Well again, the last thing that we do is the type of function. And so here the last thing we do is we divide. And so, well that's not a constant. It's not a constant multiplied by a function. It's not a sum. It's not a power. There is no rule that tells us what to do if we have a quotient. At least not yet. So what should we do now? We'd always fall back and do a little bit of algebra. This quotient can be rewritten as x cubed over x plus square root of x over x. And we'll do a little bit of algebraic simplification and cleanup. The type of function is a sum. And so we can break this apart as the sum of the derivatives. And now both terms are of the form x to the power n. So the derivatives will be. And we should do a little bit of algebraic cleanup to get our final answer. Or how about the derivative of 2x plus 5 times 3x plus 7? So again the type of function is determined by the last operation performed. And in this case we are multiplying two things. So this is a product. But we don't yet know how to differentiate a product. But we can do some algebra. So when we differentiate, we're differentiating a sum. So this becomes the derivative of 6x squared plus the derivative of 29x to the first plus the derivative of 35. Now these two, 6x squared and 29x to the first, are constant multiples of a function. So that constant multiplier, 6 or 29, can be moved outside. And now we have the derivative of x to the second, the derivative of x to the first, the derivative of a constant. And we know how to differentiate those things. And we end up with our final answer.