 Let's take a look at a couple of examples of how to use the limit definitions of the derivative to find the derivative of an equation. So let's first one, we have an equation x squared plus 1 and we are looking to find the slope of the tangent to that graph at the point negative 1, 2. Now since the point at which we want to find the slope of the tangent line, also known as the derivative, we want to use the second limit definition as that will be a little easier. So as we set it up, the C value that is mentioned in that limit definition of the derivative is your negative 1. So we are specifically trying to find the derivative at negative 1. So according to that second limit definition, it's going to be the limit as x approaches negative 1, of the function itself, f of x, minus f of negative 1, which is in this case 2, we are given that in the ordered pair. And that's going to be over x minus the C value. So in this case that becomes x plus 1. So as we go ahead and continue on from there, you really just have a limit that you want to evaluate. So we're doing the limit as x approaches negative 1. When we simplify the numerator, we have x squared minus 1, so we'll be able to factor that in a second. It's the difference of two squares, of course. The x plus 1s cancel out. So now when we substitute negative 1 in for the x here, we get negative 2. So that would be the slope of that curve at that point negative 1, 2. And we'll take a look at that graphically in just a second. But let's take a look at how we would have done this if we wanted to use the first limit definition of the derivative instead. So as we set it up, the first limit definition, remember, gives you a general equation or expression for the derivative of a given graph. So what we're finding in this case is f prime of x and that's really going to be our slope and a general expression for the slope. So this would be the limit as h approaches 0. Now remember in the numerator, we need f of x plus h. So in place of the x in our equation x square plus 1, we need to put x plus h. So when we multiply that out and square it out, we have x squared plus 2xh plus h squared. All right, so that's the x plus h quantity squared part. But remember then it was plus 1 according to the graph we were given. So that's your f of x plus h minus f of x. So that's going to be the function itself all over h. So now we just have some simplifying to do. So in the numerator when you distribute that negative over the x squared plus 1 quantity, the x-squareds cancel out and you have remaining 2xh plus h squared and that's over h. So once again, you're back to having a limit to evaluate. Obviously you cannot substitute 0 in place for h because of the denominator. So we're going to have to algebraically manipulate it somehow. So we can factor out the h from the numerator. That then allows your h's to cancel out and now you can substitute the 0 in place of h and you get 2x. So what this gives us then is a general expression for how to get the slope to this curve x squared plus 1 at any x you might substitute in. Now remember we were trying to find the slope specifically at negative 1. So if we now substitute the negative 1 in, notice how you get that same answer that we got from the second limit definition of the derivative. So let's take a look at this graphically. So I've substituted into y1, our equation x squared plus 1 and let's take a look at the graph. Obviously it's a parabola. We wanted the slope of the tangent line to the curve at the point negative 1, 2. So remember you can use some of your calculator features. You can do second trace and let's go down to number 6, dy dx, we want at negative 1 and there's your answer right there. So it's a great way to use your calculator to check your answers and if you take a look at the parabola at that side of the curve, obviously it is on a down slope, it is decreasing there. So that's why the derivative, the slope of the tangent line is coming out negative. So let's take a look at another example problem. So here we're trying to find f prime of x, the derivative for f of x equals square root of x. Notice we are not given a specific point at which we want to find that derivative, the slope of the tangent line. So in this case we're trying to find a general expression for how we could get the derivative at any point x value we might substitute in. So we are going to have to use the first limit definition of the derivative. So we have f prime of x is equal to the limit as h approaches 0. So remember it's f of x plus h, so in place of the x here we need to substitute x plus h. So that turns it into square root of x plus h minus f of x, so that would be minus square root of x all over h. Once again we have a limit that we need to algebraically evaluate. We cannot substitute 0 in because of the denominator. So think back to how you learned how to algebraically evaluate limits. This is one in which you're going to have to multiply by the conjugate. So we're going to multiply by the conjugate of that numerator. So we're going to multiply by square root of x plus h plus square root of x. Both in the numerator and denominator remember that the reasoning behind that is essentially you're really just multiplying that given expression, that rational expression by one since it's the same thing over itself. So if we multiply out in the numerator, so square root of x plus h times square root of x plus h is simply the quantity x plus h. The outsides and the insides cancel out because one quantity is positive, the other is negative, and then you're just left with minus square root of x times square root of x, so that's just minus x. In the denominator we're just going to write it out as it is. We're not actually going to distribute that h, and you'll see why in a second you might remember this from when you've seen it before in the earlier lesson on how to evaluate limits algebraically. If you take a look at your numerator, the x's will cancel out and reduce to simply h. That then cancels out with this. Now just don't forget you have a one now in the numerator. So now you're able to substitute the zero in place of the h here, so you have one over square root of x plus square root of x, so your answer in the end is one over two square root of x. So remember what this represents. This is a general expression as to how to get the slope at any point x you might substitute in. Later on, once you learn another rule in calculus, you'll be able to get to this answer a lot quicker than we did here by using the limit definition of the derivative.