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

Lesson 9: The product rule# Proving the product rule

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.(60 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?(11 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!(27 votes)

- pretty cool, would have never figured it out on my own either(23 votes)
- beautifully simple(15 votes)
- how can people think of the yellow part, that's insane(9 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) ?(5 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).(6 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!(3 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.(6 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*(4 votes)

- 6:43, I hate this and i'm beyond confused here.

when multiplied, lim h->0 multiplies both f(x+h) and the derivative (and the other half, too lazy to include that here)! That doesn't make any sense.

Take the equation a*(bd+cf). When multiplied, it becomes abd + acf, it doesn't become ab * ad + ac * af, that is not how that works. This notion implies a*(bd+cf) = ab * ad + ac * af. Maybe there is something i'm missing here, but that is simply not how multiplying works.

Second, HOW DOES THIS HELP WITH THE ACTUAL EQUATION?

Think this, lim h -> 0 defines h to be infinitely close to 0, so why on god's green earth would we "multiply" the definition of h to the entire equation? All I can see is that it re-defines the same definition to all sides of the equation.(0 votes)- The fact that you can write it like that comes from the very definition of the limit of a product of two functions, which says that the "limit of the product of two functions equals the product of their limits". If you want, you can use the epsilon-delta definition to prove this for yourself. So, that's why lim (h --> 0) of (g(x)(f(x+h)-f(x))/h) = lim(h-->0) g(x) * lim (h-->0) ((f(x+h)-f(x))/h).

"This notion implies a*(bd+cf) = ab * ad + ac * af. Maybe there is something i'm missing here, but that is simply not how multiplying works."

Where does it imply that? Here too, a*(bd+cf) implies abd + acf. Nowhere in Math will a*(bd+cf) imply ab * ad + ac * af. Not sure why you felt that way.

"Think this, lim h -> 0 defines h to be infinitely close to 0, so why on god's green earth would we "multiply" the definition of h to the entire equation? All I can see is that it re-defines the same definition to all sides of the equation."

We're not multiplying the definition of h. Basically, instead of multiplying the functions and then taking the limit as the h in the product tends to zero, we take the limit of h tending to zero for each component and then multiply the results. I think it's pretty intuitive and as I said, epsilon-delta provides a rigourous way to prove it.(8 votes)