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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition) > Unit 3

Lesson 6: Product rule# Proving the product rule

AP.CALC:

FUN‑3 (EU)

, FUN‑3.B (LO)

, FUN‑3.B.1 (EK)

Proving the product rule for derivatives.

The product rule tells us how to find the derivative of the product of two functions:

The AP Calculus course doesn't require knowing the proof of this rule, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

## Without further ado, we present to you the proof!

## Want to join the conversation?

- Is there reasons why there are many proofs?(16 votes)
- There are several proofs because each statement or concept needs to be proven individualy in order to be accepted. Proofs can also bundle up and be used to prove an even higher concept.(47 votes)

- According to the Limit rules lim x->a f(x).g(x) = lim x->a f(x) . lim x->a g(x) . But why when the derivative is calculated this rule is violated?(10 votes)
- Interesting question! The derivative of a function is not just the limit of the function, but rather is a limit of something more complicated that is related to the function.

Note that the derivative of f(x)g(x) is limit h-->0 of [f(x+h)g(x+h) - f(x)g(x)]/h,

the derivative of f(x) is limit h-->0 of [f(x+h) - f(x)]/h,

and the derivative of g(x) is limit h-->0 of [g(x+h) - g(x)]/h.

In general, [f(x+h)g(x+h) - f(x)g(x)]/h is not the product of [f(x+h) - f(x)]/h and [g(x+h) - g(x)]/h, so we can't just use the product property of limits to conclude that the derivative of f(x)g(x) is the product of the derivatives of f(x) and g(x).

Have a blessed, wonderful day!(20 votes)

- At1:00I did not understand why d/dx[f(x)*g(x)] =

lim [f(x+h)g(x+h) - f(x)g(x)]/h

h->0(6 votes)- This is the definition of the derivative with f(x)·g(x) substituted in for f(x).(7 votes)

- Why does lim h->0 f(x + h) is equal to f(x) ?(3 votes)
- Because if a function is continuous, then the limit of the function is equal to the function of the limit. That is,

lim_{x-> c} f(x)=f(lim_{x->c} x)=f(c)

In this case, we have assumed f to be differentiable, and therefore continuous, so lim_{h->0} f(x+h)=f(x+0)=f(x).(5 votes)

- How does Sal go from step 2 to step 3 (minute5:47to6:26)? I mean, how is it concluded that lim h->0 [f(x+h)*(g(x+h)-g(x))/h] = lim h->0 [f(x+h)]*lim h->0 [(g(x+h)-g(x))/h]?

In other words, if the limit of two functions multiplying each other is lim h->0 [f(x)g(x)] = f’(x)g(x) + f(x)g’(x), then why does Sal says that “the limit if the products is the product of the limits”? Wouldn’t we fall into a redundance in the proof? Please help!(2 votes)- You're confusing the product rule for derivatives with the product rule for limits. The limit as h->0 of f(x)g(x) is

[lim f(x)][lim g(x)], provided all three limits exist. f and g don't even need to*have*derivatives for this to be true.

The derivative of f(x)g(x) if f'(x)g(x)+f(x)g'(x). This is not the same as the limit.(5 votes)

- at1:50, did we assume the two functions be on a same graph or something? i did not get the step... why do we multiply (f.g)(x+h) and subract it by f.g(x) and have only one h in the denominator? how to visualise two functions of distinct graphs, f(x) and g(x) being multiplied, that too of their slopes?

thanks(2 votes)- Let the function H(x) = f(x)*g(x). H(x) can be plotted on the same graph as f(x) and g(x) because all 3 of them are functions of "x" and only "x".

We have to find the derivative of H(x), which is*lim[h-->0] (H(x+h) - H(x))/h*

H(x+h) is f(x+h)g(x+h) and H(x) is f(x)g(x), the limit can be written as*lim[h-->0] (f(x+h)*g(x+h) - f(x)*g(x))/h*(2 votes)

- According to the Limit rules lim x->a f(x).g(x) = lim x->a f(x) . lim x->a g(x) . But why when the derivative is calculated this rule is violated?(2 votes)
- The product limit rule is not violated.

(f(x)g(x))' = lim h->0 (f(x+h)g(x+h) - f(x)g(x)) / h

We cannot split the limit into the product of lim h->0 (f(x+h) - f(x))/h and lim h->0 (g(x+h) - g(x))/h.(2 votes)

- When the f(x+h) and g(x) were factored out of the numerator, why were there no parentheses? The way it was written, it seems like the f(x+h) and g(x) would just be multiplied to the numerator.(1 vote)
- That's because they are both multiplied to the numerator! ;)(3 votes)

- At1:09, why wouldn't the denominator be h^2? If you multiply the derivative formula for f(x) and the derivative formula for g(x), wouldn't the h become h^2?(2 votes)
- Let 𝑠(𝑥) = 𝑓(𝑥)𝑔(𝑥)

Then 𝑠'(𝑥) = lim(ℎ → 0) (𝑠(𝑥 + ℎ) − s(𝑥))∕ℎ =

= lim(ℎ → 0) (𝑓(𝑥 + ℎ)𝑔(𝑥 + ℎ) − 𝑓(𝑥)𝑔(𝑥))∕ℎ(1 vote)

- At8:05and8:42, I understand that "lim as h approaches 0, (g(x+h)-g(x))/h" and "lim as h approaches 0, (f(x+h)-f(x))/h" are g'(x) and f'(x) respectively, but if zero is plugged into h, aren't they "undefined" since their denominators are zero?(1 vote)
- That is why we take the limit. The limit is still defined even though the expression is undefined at h=0. This is one of the main reasons limits are useful. If you are still confused, you should probably review the unit on limits.(1 vote)