In some nice cases, you can calculate the derivative of a function directly from the definition of derivative. Here's an example. Suppose I've got a function f of x equals x squared. I want to calculate this derivative. Well, the derivative of this function at x is the limit as h goes to zero of f of x plus h minus f of x over h. I know what f is, so I can just plug that in. This is the limit as h goes to zero of what's f of x plus h. I plug an x plus h for x and I get x plus h squared minus f of x, which is just x squared over h. I can calculate that limit. That's the limit of x plus h squared minus x squared. Let's expand out the numerator, right? This is the limit as h goes to zero of what's x plus h squared, it's x squared plus 2xh plus h squared minus x squared over the denominator h. Now, in this situation, right, I've got an x squared and a minus x squared, so those terms die. I want to my left with just 2xh plus h squared over h. I'm taking the limit as h goes to zero. I'm going to factor out an h from the numerator. This is the limit as h goes to zero of what? I'll have h times 2x plus h. Here, I've factored out the h from the numerator divided by h. I'm taking the limit as h goes to zero of h times 2x plus h over h. This is not defined when h equals zero, right? But this thing has the same values as just 2x plus h as long as h is in zero, so they have the same limit. This is the same as the limit as h goes to zero of just 2x plus h. Another way to see this, right, when people just cancel these terms, right? I'm going to cancel those h's. I've got the limit as h goes to zero of this function, which isn't defined at zero, but it's value is away from zero or the same as the values of 2x plus h, so it's got the same limit as just 2x plus h. Now, how do I take the limit of 2x plus h? That's the limit of a sum, and the limit of a sum is the sum of the limits. So this is the same as the limit of 2x as h goes to zero plus the limit of h as h goes to zero as long as these limits exist. But they do, right? The limit of 2x as h goes to zero, this is the limit of a constant, right? 2x doesn't have any h's in it. It's constant as far as h is concerned, right? h is the thing that's moving here, and 2x isn't moving when h is moving. So this is just 2x. So it's a limit of something that's not changing. So this is 2x. And once the limit of h as h goes to zero, just zero, right? So this is 2x plus zero. I mean, just it's just 2x, right? And that is the derivative of x squared, right? But we calculated it just by using the definition of derivative and all the properties that we know for limits. There's other examples that you could try to do this way too, right? I mean, you could try to do x cubed, for example, this way. And, you know, in general, x to the n.