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## AP®︎/College Calculus BC

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

Lesson 12: Optional videos- Proof: Differentiability implies continuity
- Justifying the power rule
- Proof of power rule for positive integer powers
- Proof of power rule for square root function
- Limit of sin(x)/x as x approaches 0
- Limit of (1-cos(x))/x as x approaches 0
- Proof of the derivative of sin(x)
- Proof of the derivative of cos(x)
- Product rule proof

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# Product rule proof

Let's delve into the proof of the product rule, a key concept in calculus. We apply the definition of a derivative to the product of two functions, making sense of this rule. Through smart algebraic manipulation, we arrive at the classic product rule formula.

## Want to join the conversation?

- how can we say that d/dx (f(x).g(x)) = lim(h --> 0) [f(x+h).g(x+h) - f(x).g(x)]/h

did not understand about the numerator of the limit or the derivative.(14 votes)- A good, formal definition of a derivative is, given
`f(x)`

then`f′(x) = lim(h->0)[ (f(x-h)-f(x))/h ]`

which is the same as saying if`y = f(x)`

then`f′(x) = dy/dx`

.`dy = f(x-h)-f(x)`

and`dx = h`

. Since we want h to be 0,`dy/dx = 0/0`

, so you have to take the limit as h approaches 0.

So now, we're starting with a particular form of f(x) where`f(x) = g(x)*k(x)`

. So, we plug g(x)*k(x) into our original equation to figure out its derivative.`f′(x) = lim(h->0)[ (f(x-h)-f(x))/h ]`

`f′(x) = lim(h->0)[ (g(x-h)*k(x-h) - g(x)*k(x)) / h ]`

(27 votes)

- The little trick he uses is not sufficient enough for me to prove to myself the product rule. Seems like a shot in the blind. Thankfully we already know what the product rule is so we sort of know what we need to add the expression but there has to be a more intuitive way to prove the product rule? Also he's multiplying the entire derivations of 2 functions so why isn't it ((f(x+h)-f(x))*(g(x+h)-g(x)))/h ?????? Am guessing because the product of 2 derived functions is not the same as derivation of products of functions? This would mean that f(x)*g(x) is now one new function, it's not 2 seperate functions but one as in multiplications where ab is not 2 terms but 1 term, so f(x)*g(x)=F(x), if we evaluate F(x) we get (F(x+h)-F(x))/h and like I said, F(x)=f(x)*g(x). So replace F(x) and F(x+h) with corresponding terms and you get what Sal got, did I get this right? I really do need to get this right otherwise my brain will refuse to memorize it. That is, it's easier for me to understand something instead memorizing the entire thing. That way I can always a problem down into what I know and therefore I can always find the formula for whatever problem I might have. Also watching Implicit differentiation and logarithmic differentiation has led me to believe this could be proven with natural logarithm but how?(8 votes)
- It is not a shot in the blind at all. It is a clever use of the fact that adding and subtracting the same value does not change an expression. The ability to see that such a device can work is the key to many proofs in mathematics, and it is this type of insight that separates the mathematicians from those that just do math without thinking about what it means. It is this type of insight and intuition, that being, the ability to leverage the rules of mathematics creatively that produces much of the beauty in math.

I think you do understand Sal's (AKA the most common) proof of the product rule.

Having said that, YES, you can use implicit and logarithmic differentiation to do an alternative proof:

y=f(x)g(x)

ln(y) = ln (f(x)g(x)) = ln(f(x)) + ln(g(x))

Take the derivative of both sides:

y'/y = f'(x)/f(x) + g'(x)/g(x)

Solve for y'

y' = y(f'(x)/f(x) + g'(x)/g(x))

Since y = f(x)g(x)

y' = f(x)g(x)(f'(x)/f(x) + g'(x)/g(x))

Multiply it all out and you get

y' = g(x)f'(x)+f(x)g'(x)

Since y' = d/dx[y] and y = f(x)g(x) we have

d/dx[f(x)g(x)] = g(x)f'(x) + f(x)g'(x).(36 votes)

- at5:43, why is the limit of the product the product of the limits? Have any previous videos discussed this?(10 votes)
- Some limit properties are presented here

https://www.khanacademy.org/math/differential-calculus/limits-topic/algebraic-limits/v/limit-properties

but they're not proved here; it seems like there aren't any khanacademy videos on the proofs of these properties.(5 votes)

- There are other ways to do it? This way seems pretty straight forward; how can you prove it any other way?(5 votes)
- Let y=v.u, where v and u are functions in x

so y+δy=(v+δv)(u+δu) when we let x change by a small amount

if we expand the brackets we get

y+δy= v.u + u.δv + v.δu + δv.δu

as y=v.u,we can eliminate v.u from the right and y from the left

we then divide by δx to get δy/δx= u .δv/δx +v .δu/δx +(δv.δu)/δx

as δx tends to 0, (δv.δu)/δx becomes insignificant

so,we end up with dy/dx=u .dv/dx + v .du/dx,

which is another way of writing the product rule(12 votes)

- How is the
`lim h->0 f(x+h)`

the same as`f(x)`

? that wouldnt be true for a discontinuous function. the`limit h->0 f(0+h)`

is infinity but f(0) is not defined. So those expressions are not equal to each other.(5 votes)- the derivative is only defined where a function is defined. Going a bit farther you will see that the derivative (which is a limit) can only exist if the function is continuous at that point. If the function is not defined at a you certainly cannot take the derivative at x=a, since a isn't in the domain of the function and you cannot set up a different quotient at a.

We often say "differentiability implies continuity" since to have a derivative at a point a function must be continuous there (not just defined there). Even if it is continuous, it still might not be differentiable--think of the absolute value function at x=0. So continuity is a necessary but not sufficient condition for differentiability.(6 votes)

- At1:45why doesnt he do FOIL?(5 votes)
- No, that's not right.

You could do that with (x + h) (x + h)

but he has f(x + h) * g(x + h)

Which is entirely different.(6 votes)

- at2:46sal left a space and added and subtracted something to get in the required form but what about the real things which we get after multiplying {f(x+h)-f(x)/h}X {g(x+h)-g(x)/h}(4 votes)
- This is just very difficult for us to do; as Sal said, at this point, there would be no easy way to work with that expression. The method he used was a little 'trick' to get around that and make the expression a little more manageable. He didn't change the original expression, however, because he added AND subtracted the new part.(2 votes)

- from where did sal come up with the first equation written in white..........

i know i saw it somewhere at khan academy jst help me find that video(2 votes) - So if we can claim that the the limit as h→0 of f(x+h) is f(x), as was stated in the video at7:30, Why can't you evaluate it as f(x+h)g(x+h)-f(x)g(x)/h = f(x)g(x+h)-f(x)g(x)/h = f(x)((g(x+h)-g(x))/h), which would be equal to f(x)g'(x). This result is clearly wrong, but I can't see where exactly I've made a mistake.(3 votes)
- You need to put the entirety of the first expression in parentheses as it all must be divided by h(2 votes)

- I've seen this metod of adding and subtracting (or multiplying and dividing) a lot of times.

sine of sum and chain rule {*\}.

a^2-b^2, product rule and directional derivative {+-}.

It seems that we are solving the same kind of problem over and over again. Is there any generalization of this method in terms of more abstract things?(3 votes)

## Video transcript

- [Voiceover] What I
hope to do in this video is give you a satisfying
proof of the product rule. So let's just start with our
definition of a derivative. So if I have the function F of X, and if I wanted to take
the derivative of it, by definition, by definition, the derivative of F of X is the limit as H approaches zero, of F of X plus H minus F of X, all of that over, all of that over H. If we want to think of it visually, this is the slope of the
tangent line and all of that, but now I want to do something a little bit more interesting. I want to find the derivative with respect to X, not just of F of X, but the product of two functions, F of X times G of X. And if I can come up with
a simple thing for this, that essentially is the product rule. Well if we just apply the
definition of a derivative, that means I'm gonna take the limit as H approaches zero, and the denominator I'm gonna have at H, and the denominator,
I'm gonna write a big, it's gonna be a big rational expression, in the denominator I'm gonna have an H. And then I'm gonna evaluate
this thing at X plus H. So that's going to be F of X plus H, G of X plus H and from
that I'm gonna subtract this thing evaluated F of X. Or, sorry, this thing evaluated X. So that's gonna be F of X times G of X. And I'm gonna put a
big, awkward space here and you're gonna see why in a second. So if I just, if I evaluate this at X, this is gonna be minus F of X, G of X. All I did so far is I just
applied the definition of the derivative, instead
of applying it to F of X, I applied it to F of X times G of X. So you have F of X plus H, G of X plus H minus F of X, G of X, all of that over H. Limit as H approaches zero. Now why did I put this
big, awkward space here? Because just the way I've
written it write now, it doesn't seem easy to
algebraically manipulate. I don't know how to evaluate this limit, there doesn't seem to be
anything obvious to do. And what I'm about to show you, I guess you could view it
as a little bit of a trick. I can't claim that I would
have figured it out on my own. Maybe eventually if I
were spending hours on it. I'm assuming somebody was
fumbling with it long enough that said, "Oh wait, wait. "Look, if I just add and
subtract at the same term here, "I can begin to
algebraically manipulate it "and get it to what we all know "as the classic product rule." So what do I add and subtract here? Well let me give you a clue. So if we have plus, actually, let me change this, minus F of X plus H, G of X, I can't just subtract, if I subtract it I've got to add it too, so
I don't change the value of this expression. So plus F of X plus H, G of X. Now I haven't changed the value, I just added and
subtracted the same thing, but now this thing can be manipulated in interesting algebraic ways to get us to what we all love about the product rule. And at any point you get inspired, I encourage you to pause this video. Well to keep going, let's just keep exploring this expression. So all of this is going to be equal to, it's all going to be equal to the limit as H approaches zero. So the first thing I'm gonna do is I'm gonna look at, I'm
gonna look at this part, this part of the expression. And in particular,
let's see, I am going to factor out an F of X plus H. So if you factor out an F of X plus H, this part right over here is going to be F of X plus H, F of X plus H, times you're going to be left
with G of X plus H. G of, that's a slightly
different shade of green, G of X plus H, that's that there, minus G of X, minus G of X, oops, I forgot the parentheses. Oops, it's a different color. I got a new software program
and it's making it hard for me to change colors. My apologies, this is not
a straightforward proof and the least I could do is
change colors more smoothly. Alright, (laughing) G of X plus H minus G of X, that's that
one right over there, and then all of that over this H. All of that over H. So that's this part here and then this part over here this part over here, and
actually it's still over H, so let me actually circle it like this. So this part over here I can write as. So then we're going to have plus... actually here let me, let me factor out a G of X here. So plus G of X plus G of X times this F of X plus H. Times F of X plus H minus this F of X. Minus that F of X. All of that over H. All of that over H. Now we know from our limit properties, the limit of all of this business, well that's just going
to be the same thing as the limit of this as H approaches zero plus the limit of this
as H approaches zero. And then the limit of the product is going to be the same thing
as the product of the limits. So if I used both of
those limit properties, I can rewrite this whole
thing as the limit, let me give myself some real estate, the limit as H approaches
zero of F of X plus H, of F of X plus H times, times the limit as H approaches zero,
of all of this business, G of X plus H minus G of X, minus G of X, all of that over H, I think you might see where this is going. Very exciting. Plus, plus the limit, let me write that a
little bit more clearly. Plus the limit as H
approaches zero of G of X, our nice brown colored G of X, times, now that we have our product here, the limit, the limit as H approaches zero of F of X plus H. Of F of X plus H minus F of X, minus F of X, all of that, all of that over H. And let me put the parentheses
where they're appropriate. So that, that, that, that. And all I did here, the limit, the limit of this sum, that's gonna be the sum of the limits, that's gonna be the limit of this plus the limit of that, and then the limit of
the products is gonna be the same thing as the
product of the limits. So I just used those
limit properties here. But now let's evaluate them. What's the limit, and I'll do them in different colors, what's this thing right over here? The limit is H approaches
zero of F of X plus H. Well that's just going to be F of X. Now, this is the exciting part, what is this? The limit is H approaches
zero of G of X plus H minus G of X over H. Well that's just our, that's the definition of our derivative. That's the derivative of G. So this is going to be, this is going to be the
derivative of G of X, which is going to be G prime of X. G prime of X. So you're multiplying these two and then you're going to have plus, what's the limit of H
approaches zero of G of X? Well there's not even any H in here, so this is just going to be G of X. So plus G of X times the limit, so let's see, this one is in brown, and the last one I'll do in yellow. Times the limit as H approaches zero, and we're getting very close, the drum roll should be starting, limit is H approaches zero of F of X plus H minus F of X over H. Well that's the definition
of the derivative of F of X. This is F prime of X. Times F prime of X. So there you have it. The derivative of F of
X times G of X is this. And if I wanted to write
it a little bit more condensed form, it is equal to, it is equal to F of X times the derivative
of G with respect to X times the derivative
of G with respect to X plus G of X, plus G of X times the derivative
of F with respect to X. F with respect to X. Or another way to think about it, this is the first function
times the derivative of the second plus the second function times the derivative of the first. This is the proof, or a proof, there's actually others
of the product rule.