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

Lesson 1: Basic differentiation rules- Proof of the constant derivative rule
- Proofs of the constant multiple and sum/difference derivative rules
- Basic derivative rules: find the error
- Basic derivative rules: find the error
- Basic derivative rules: table
- Basic derivative rules: table
- Basic differentiation review

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# Proofs of the constant multiple and sum/difference derivative rules

Constant multiple rule says that the derivative of f(x)=kg(x) is f'(x)=kg'(x). Sum/Difference rule says that the derivative of f(x)=g(x)±h(x) is f'(x)=g'(x)±h'(x). Sal introduces and justifies these rules.

## Want to join the conversation?

- @1:35where did you get the h from?(7 votes)
- We get that h from the definition of a derivative:

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

'h' basically represents a very little increase in x value.

instead of 'h' some book also have (delta x) [delta looks like a small triangle, its a greek symbol](12 votes)

- At0:50Sal chooses algebraic argument over graphical one, but in the end boils it down to limits' properties, which weren't proved here, and given as "common sense", which is okay with me.. but maybe there is a more thorough proof that includes limits' properties?(4 votes)
- The use and proofs of limits started here:

https://www.khanacademy.org/math/differential-calculus/limit-basics-dc#formal-definition-of-limits-dc

which was then developed and used to provide the limit definition of the derivative here:

https://www.khanacademy.org/math/differential-calculus/derivative-intro-dc/formal-definition-of-derivative-dc/v/alternate-form-of-the-derivative

So by this video, the limit definition of the derivative is proven and we use the limit definition of the derivative to prove the assertions made here, ie, d/dx[f(x)=kg(x)] is f'(x)=kg'(x) and d/dx[f(x)=g(x)±h(x)] is f'(x)=g'(x)±h'(x).

Using the limit definition of the derivative (@1:43) I highly recommend you attempt to prove these assertions yourself.

Cheers!(5 votes)

- So can dx1/dt - dx2/dt = d(x1- x2) / dt ? PLEASE HELP(2 votes)
- Yes, although a more typical notation would be something like
`dx1/dt - dx2/dt = d/dt [ x1 - x2 ]`

.

We often use the d/dt (or d/dx, etc.) as a kind of operator to indicate that we're taking the derivative of whatever function comes after the d/dt.

Hope that helps.(6 votes)

- random question over rigorous-ness in mathematics. How do we know that 1+1=2?(1 vote)
- We know it from an area of mathematics known as Foundation of Mathematics. This is going to sound paradoxical, but actually
*proving*that 1 + 1 = 2 is way above the levels of mathematics covered on this site. If I were you, I'd take the basic laws of arithmetic as axiomatic - for now at least. If you're interested, there is a proof in Russell and Whitehead's*Principia Mathematica*. Be warned: it is several*hundred*pages of proof!(5 votes)

- At2:13, shouldn't it be (k(g(x+h)-g(x))) / h, since there's no k in the denominator I don't understand how you get k((g(x+h)-g(x)) / h)?(2 votes)
- k does not have to be in the denominator to factor it out. For example, 2(x/5) = 2/1(x/5) sense 2= 2/1 this can simplified to 2x/5. Conversely 2x/5 can be broken down to 2/1(x/5) or 2(x/5) note: 2 was not in the denominator this is not breaking any rules of math. https://www.khanacademy.org/math/arithmetic/fraction-arithmetic/arith-review-multiply-fractions/v/multiplying-fractions . Hopefully this will help you(2 votes)

- Does differentiable mean the derivatives of the 2 functions ARE the same or NOT the same?

For example in this question my teacher gave us a hint that we would need to find the left & right hand derivatives:

Is f differentable at x=1?

Let f(x)= 2-x if x</= 1

and (x^2)-2x+2 if x>1(1 vote) - how would you write rule 2 f(x)= g(x)+j(x) => f'(x) g'(x)+j'(x) in Leibniz notation?(0 votes)
- @0:48shouldn't f'(x)=0?cause when f(x)=k,f'(x)=0.Which means f'(x)=0*g(x).So when we multiply anything by zero is always zero right?(1 vote)
- We have f(x)=k•g(x). Take the derivative of both sides to get f'(x)=d/dx (k•g(x)). So what does the right hand side simplify to?

The derivative of a product is*not*the product of the derivatives. That is, it's not the case that d/dx(f(x)g(x))=f'(x)g'(x). If that were the case, then every derivative would be 0, since g(x)=1•g(x). That's not useful.

Sal goes on to prove in the video why the constant gets moved outside the derivative.(2 votes)

- Is it just me, who's lost and confused on whats happening and don't know where to start?(1 vote)
- It's very unlikely that it's just you.

Try watching the video again slowly. Pinpoint the exact sentence where you started feeling lost, then ask a question or go back to a previous video to try to understand that sentence.(2 votes)

- How does f(x) = k * g(x) -> f'(x) = k g'(x) lead to something like f'(3x) = 3 f'(x). I was able to do the practice exercise but I don't really feel I understand it well.(1 vote)
- It doesn't imply that, and f'(3x) doesn't necessarily equal 3f'(x). As a counterexample, let f(x)=x³. Then f'(3x)=27x², while 3f'(x)=9x².

What it would imply is that the derivative of 3f(x) is 3f'(x).(2 votes)

## Video transcript

- [Voiceover] In the last
video we introduced you to the derivative property right over here that if my function is
equal to some constant that the derivative is going to be zero at any X, and we made a graphical argument and we also used the
definition of the limits to feel good about that. Now let's give a few
more of these properties and these are core properties as you throughout the rest of your calculus life your career, you will be
using some combination of these properties to find derivatives. So it's good to, one, know about them, and to feel good that
they're actually true. The second one is if my function F of X is equal to some constant times another function G of X well, then the derivative of F of X is going to be equal to that same constant. Let me do it, that same constant times the derivative of the G of X. And once again, maybe
we can make a actually we could make a graphical
argument for why that is true. This is going to multiply the slope is one way to think about it. But it's easier to make
an algebraic argument just using, frankly we can
use either one of these definitions of the derivative. I'll use the one on the right because it feels more general although you can say,
well this is true for A, but X could be any A. But I'll just use the one on the right. So, if we want to find F prime of X, F prime of X using this definition, we know, we know. F prime of X is going
to be equal to the limit as X, I'm sorry, as H approaches zero. I'm using that definition. Of F of X plus H minus F of X all of that over H. Well, what is F of X plus H? This is the limit as H approaches zero. F of X plus H is K times G of X plus H. Minus F of X, well that's just KG of X. KG of X. All of that over H. And then you can factor the K out. This is going to be equal to the limit as H approaches zero of K times G of X plus H minus G of X all of that over H. All I did was factor that K out. And we know from our limit properties that this is the same thing as K times the limit as H approaches zero of G of X plus H minus G of X, all of that over H. And of course, all of this
business right over there. That is just G prime of X. So, this is equal to K times G prime of X. And I know what you might be thinking. Whoa, hey this feels like it was probably going to be true, so I
just assumed it was true. But you can't just assume that. I don't know, sometimes you can when you're first trying
to get your head around it you can say, oh, this seems
like a reasonable thing. But in math, we like to
really know that it is true otherwise we will build all sorts of conclusions based on unsound foundations. This allows us to ensure
that this is something that we can do. So, this is, it's good to go through what might feel like a little bit of work to
get to this conclusion. Now let's do the third property. The third property is the idea that if I have some function that's the sum or the difference of two other functions, so G of X, and let's see, I'm using H a lot so, H, let's say, I don't know, J of X. J, yeah sure, why not, J of X. You don't see a lot of J of X's out there. Well then, well then, F prime of X is going to be equal to G prime of X plus J prime of X. And this would also have been true if this is, instead of
it being a positive here, if this was a negative. Instead of addition, if
this was subtraction. If the sum were a
difference of two functions then your derivative is gonna be the sum or the difference of their derivatives. And once again, we can
just go to the definition of F prime of X. So, F prime of X is going to be equal to the limit as H approaches zero of F of X plus H. But what is F of X plus H? Well, that's G of X plus H plus J of X plus H. So that's F of X plus H minus F of X. So F of X is G of X. G of X plus J of X plus J of X. Notice, this is F of X plus H minus F of X and we're gonna put all of that over H. So we can put all of that all of that over H. Well, what is that equal to? Well, we could just rearrange what we see on top here. This is equal to the limit as H approaches zero. Well let's see, all the mentions of G of X I'm gonna put up front. So G of X plus H minus G of X. Plus J of X plus H minus J of X. And then all of that, I could write it as this, all of that over H. Or, I could, that's the same thing as this over H plus that over H. And once again, we know from our limit properties that that is the exact same thing as the limit as H approaches zero, of G of X plus H minus G of X all of that over H plus the limit as H approaches zero of J of X plus H minus J of X all of that over H. And this, right over here, that is the definition of G prime of X. And this right over here is J prime, J prime of X and we're done. And this, instead of a positive, this was instead of addition this was subtraction. Well then that subtraction
would carry through and then instead of addition here we would have subtraction. So hopefully this makes you feel good about these properties. The properties themselves
are somewhat straightforward. You could probably guess at them. But it's nice to use the definition of our derivatives to actually feel that they are very
good conclusions to make.