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

Lesson 9: The product rule

# Worked example: Product rule with table

We explore how to evaluate the derivative of the product of two functions, f(x) and h(x), at x=3, using given values for f, h, and their derivatives. Applying the product rule, we calculate the derivative of f(x)⋅h(x) at x=3 with ease.

## Want to join the conversation?

• why can't you take the derivative of the value of the function at a point?
• 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
• 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?
• 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)
• Why did Sal make up a new derivative, g'(x)? Why didn't he just find the answer to d/dx?
• 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
• 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)].