 As with limits, knowing a few properties of the derivative will help us to find many derivatives. Let's begin by finding a simple derivative. Suppose f of x is 5. We'll use the definition to find the derivative at 7. So again, we know we should be using the definition because the problem asks us to use the definition. So we'll pull that in. And to use the derivative, we need to find f of 7 and f of 7 plus h. So we find those values. And then we substitute them into our difference quotient. And by definition, we get a derivative of 0. Now, what would happen if f of x were equal to a different number? Say, if f of x were equal to 11 or pi or square root of 7. It should be clear that regardless of the value that f of x is equal to, nothing important changes here and the derivative is always going to be equal to 0. And so this suggests an important result. If f of x equals c, a constant, then the derivative is equal to 0. Now before we go on, a quick reminder of how we write functions in algebra. If I have two functions f and g, and I want to talk about the sum f of x plus g of x, a standard compact way of writing this is as f plus g of x. Likewise, if I want to talk about the product f of x times g of x, I can write this as fg of x, their quotient f over g of x, and the constant times a function as c f of x. So I suppose I know the derivative of f and g at a certain point. Can I find the derivative of f plus g? That is to say the derivative of the function that is the sum of f of x and g of x. So a good starting point is to ask yourself, self, what do I know? And since we know f prime of 3 and g prime of 3, then we know from the definition of the derivative that both of these are equal to the limit of a difference quotient. And so while we might not know very much else, we can at the very least substitute in those different quotients. Next, what we want is to find the derivative of f plus g evaluated at 3. And so bringing back our definition, we want to evaluate f plus g of 3 plus h and f of g of 3. But again, f plus g of 3 plus h is our notation for f of 3 plus h plus g of 3 plus h. And likewise, f plus g of 3 is our notation for f of 3 plus g of 3. Now I can rearrange these terms a little bit so that I have all of the f terms in one fraction and all of the g terms in the other. And because this is the limit of a sum, then I know that I can find this limit by finding the two limits and adding them together. But wait, there's still more. Bringing back our definition of the derivative, we see that this first term is actually the limit form of the derivative of f of 3. Likewise, the second form is the limit form of the derivative of g of x at 3. And so what we have is f prime of 3 plus g prime of 3. And fortunately, I already know what those are. And so that's going to be 7. And following the chain back to its starting point, that is the value of f plus g prime of 3. And so this suggests another important derivative theorem. If f and g are differentiable at x equal to a, then the derivative of f plus g of a is f prime of a plus g prime of a. Or we might say that the derivative of a sum is the sum of the derivatives. Let's take a look at a different case. Suppose g prime of 5 is equal to 7. Find the derivative of 5g at 5. So remember that in function notation, this notation 5g means 5 times the function value. So let's pull in our definition of the derivative. So our function is 5 times g of x. So we'll substitute those into our definition. And we'll do a little bit of algebra. Now we have constant times a, a whole mess of stuff, and one of our limit theorems says that the limit of a constant times a function is the constant times the limit of the function. And so this constant multiplier can be moved to the front to get us a new limit. This expression is just the derivative at 5, and we already know that value, which means we can substitute in. And following our chain of deduction, we see that the derivative of 5g at 5 is going to be 35. And this suggests another important theorem. Let f be differentiable at a, and let c be a constant. Then the derivative of constant times function is equal to constant times the derivative of function. So let's put a few things together. First, since the problem asks us to find the derivative according to the definition, we'll pull in our definition. And the definition requires that we find f of x plus h and f of x. So finding these two expressions, then substituting them into our definition and doing a little bit of algebra, we find the derivative is equal to 2x. What about the derivative of 8x squared plus 7? It's very useful to keep in mind the following idea. The type of function is determined by the last operation performed. So here we have this expression 8x squared plus 7, and the last thing that we are supposed to do when evaluating this expression is add. And so this function is a sum, and so the first thing we ought to ask ourselves is what do we do when we want to find the derivative of a sum? And we have our result that the derivative of a sum is equal to the sum of the derivatives. So the derivative of 8x squared plus 7 is the derivative of 8x squared plus the derivative of 7. But wait, there's more. Here we want to take the derivative of 8x squared, and again we ask ourselves what's the last thing that we do? And in this case, the last thing that we do is multiply by 8. So this is a constant times a function. And so what we know is that the derivative of a constant times a function is the constant times the derivative of the function. And so we can move that 8 out front. And now we have to find two derivatives, the derivative of x squared and the derivative of 7. Well, we just figured out the derivative of x squared. So what's the derivative of 7? Since 7 is a constant, the derivative is going to be 0. And we can do a little bit of algebra to find our derivative.