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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 2

Lesson 9: The product rule# Worked example: Product rule with table

AP.CALC:

FUN‑3 (EU)

, FUN‑3.B (LO)

, FUN‑3.B.1 (EK)

Given the values of f and h (and their derivatives) at x=3, Sal evaluates the derivative of f(x)⋅h(x) at x=3.

## Want to join the conversation?

- why can't you take the derivative of the value of the function at a point?(5 votes)
- The reason we don't take the derivative of a function after evaluating it at a point is simple - after you find the value of a function at some point, what do you get? No matter what (real-valued) function you have, if you evaluate it at some specific point you will always get some plain old constant number. Since we know that the derivative of any constant number is 0, this method of differentiation will simply give you 0 no matter what function you start with.

If you derive the entire function first, then what do you have? You have a new function, that, when you plug a value into it, it gives you the slope of the original function at that point. So evaluating first then deriving -> always get 0, but deriving first and then evaluating -> finds slope of original function, which is what we want differentiation to do.

Hope this helps(21 votes)

- Wouldn't this have just been the same as multiplying f'(3) x h'(3)? Why did you use the power rule to get the answer?(3 votes)
- No, the rule for the derivative of a product is not the same as the rule for a sum/difference.

d/dx [f(x) + h(x)] = f'(x) + h'(x)

d/dx [f(x) - h(x)] = f'(x) - h'(x)

Whereas

d/dx [f(x)*h(x)] = f'(x)*h(x) + f(x)*h'(x)(10 votes)

- Why did Sal make up a new derivative, g'(x)? Why didn't he just find the answer to d/dx?(3 votes)
- He didn't need to, but he likely did it so viewers can see it differently. Sometimes problems don't use the d/dx notation(3 votes)

- But isn't the derivative of a constant always equal to 0?(1 vote)
- The derivative of a constant is indeed 0.

However, 𝑔'(𝑥) = 𝑑∕𝑑𝑥[𝑓(𝑥)⋅ℎ(𝑥)] does*not*imply that

𝑔'(3) = 𝑑∕𝑑𝑥[𝑓(3)⋅ℎ(3)].(2 votes)

- Someone please explain to me, why don't we just use

d/dx [f(x)*h(x)] = f'(x)*h'(x) instead of

d/dx [f(x)*h(x)] = f'(x)*h(x) + f(x)*h'(x)

What is the difference and why should we use the product rule?(1 vote)- You can see why this is not true in most cases (I say "most cases" because there are a
*few*functions for which the derivative of their products is equal to the product of their derivatives) by graphing . Take two functions f(x) = sin(x) and g(x) = cos(x). Let's take the point x = π/2. Now, the slope of sin(x) at π/2 is 0. And, the slope of cos(x) at π/2 is -1. So, by this analogy, the slope of f(x) * g(x) = sin(x)cos(x) must be -1 * 0 = 0. However, if you graph out sin(x)cos(x), you'll see that the slope at π/2 is equal to -1, not 0. So, the derivative of the product of two functions is not equal to the product of the derivatives. Additionally, as the product rule was derived using the First Principle, it is mathematically correct and hence, should be the one put to use.(1 vote)

- Can you view this at multiplying the matrix [f'(x), f(x)] and (vertical) [h(x), h'(x)]?(1 vote)

## Video transcript

- [Voiceover] The following
tables lists the values of functions f and h,
and of their derivatives, f prime and h prime for
x is equal to three. So all this is telling us, with x is equal to three, the value of the function
is six, f of three is six, you could view it that way. h of three is zero, f
prime of three is six, and h prime of three is four. And now they want us to
evaluate the derivative with respect to x of the
product of f of x and h of x when x is equal to three. One way you could view this is, if we viewed some function, if we viewed some function g of x. g of x as being equal
to the product of f of x and h of x, this expression is the
derivative of g of x. So we could write g prime of
x is equal to the derivative with respect to x of f of x times h of x. Which is what we see right here, and which is what we want to
evaluate at x equals three. So we essentially want to evaluate g prime of three. This is what they're asking us to do. Well, to do that, let's go first up here. Let's just think about what this is doing. They're asking us to take the derivative with respect to x of the
product of two functions that we have some information about. Well, if we're taking the
derivative of the product of two functions, you
could imagine that the product rule could prove useful here. So I'm just gonna
restate the product rule. This is going to be
equal to the derivative of the first function, f prime of x, times the second function,
not taking its derivative. Plus the first function,
not taking its derivative, f of x, times the derivative
of the second function, h prime of x. So if you're trying to
find g prime of three, well that's just going to be f prime of three times h of three plus f of three times h prime of three. And lucky for us, they give
us what all these things evaluate to. f prime of three, right
over here, they tell us. f prime when x is equal
to three is equal to six. So this right over here is six. h of three, they give us that, too. h of three, when x is
three, the value of our function is zero. So this is zero. So this first term is you
just get six times zero, which is gonna be zero,
but we'll get to that. Now f of three. f of three. Well, the function when
x is equal to three, f of three is equal to six. So that is six. And then finally, h prime evaluated at three, h prime of x when x is equal to three, h prime of x is equal to four. Or you could say this is h prime of three. So this is four. And so there you have it. This is going to evaluate
to six times zero, which, that's all just gonna be zero, plus six times four, which is going to be equal to 24. And we're done.