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# Proving the chain rule

Proving the chain rule for derivatives.
The chain rule tells us how to find the derivative of a composite function:
$\frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]={f}^{\prime }\left(g\left(x\right)\right){g}^{\prime }\left(x\right)$
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.

## First, we would like to prove two smaller claims that we are going to use in our proof of the chain rule.

(Claims that are used within a proof are often called lemmas.)

### 1. If a function is differentiable, then it is also continuous.

Proof: Differentiability implies continuitySee video transcript

### 2. If function $u$‍  is continuous at $x$‍ , then $\mathrm{\Delta }u\to 0$‍  as $\mathrm{\Delta }x\to 0$‍ .

If function u is continuous at x, then Δu→0 as Δx→0 See video transcript

## Now we are ready to prove the chain rule!

Chain rule proofSee video transcript

## Bonus: We can use the chain rule and the product rule to prove the quotient rule.

The quotient rule tells us how to find the derivative of a quotient:
$\begin{array}{rl}\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]& =\frac{\frac{d}{dx}\left[f\left(x\right)\right]\cdot g\left(x\right)-f\left(x\right)\cdot \frac{d}{dx}\left[g\left(x\right)\right]}{\left[g\left(x\right){\right]}^{2}}\\ \\ \\ & =\frac{{f}^{\prime }\left(x\right)g\left(x\right)-f\left(x\right){g}^{\prime }\left(x\right)}{\left[g\left(x\right){\right]}^{2}}\end{array}$
Quotient rule from product & chain rulesSee video transcript

## Want to join the conversation?

• In the proof of differentiability implies continuity, you separate the limits saying that the limit of the products is the same as the product of the limits. But the limit of x*1/x at zero cannot be divided as the limit of x times the limit of 1/x as the latter one does not exist.
I do understand that this works in this case because both the limits exist.
• Good! You have found where the theorem fails: that is the limit of the product is the product of the limit remains true only when both limits actually exist! Clearly if one of the limits fails to exist then any equality would be meaningless as you have argued above!!
• In the chain rule proof, at what point did you use the lemma 'differentiability implies continuity' (that you proved in a previous video)?
• He uses it when using the second lemma because the second lemma assumes the continuity.
(1 vote)
• In video 1, why is it okay to divide by (x-c)? Isn't that 0 when x -> c?
• Only if x = c. If it's approaching, then it's not quite equal yet.
• I didn't understand why if function u is continuous at x, then Δu→0 as Δx→0.
Can't a continuous function u(x) have a vertical line somewhere in its graph? In the graph of a vertical line, as the change in x approaches 0, the change in u(x) approaches infinity.
(1 vote)
• A function cannot contain a vertical line in its graph because a vertical line would imply that one input value is being mapped to multiple output values (which would violate the definition of a function).
• In the proof of the chain rule by multiplying delta u by delta y over delta x it assumes that delta u is nonzero when it is possible for delta u to be 0 (if for example u(x) =2 then the derivative of u at x would be 0) and then delta y over delta u would be undefined? This would be multiplying by 0 as we never proved that delta u was nonzero and only that it was approaching 0 in the proof above the chain ule proof. The limit of delta u as delta x approaches 0 is still 0 is delta u =0 but we then couldn’t multiply by delta u. How is it allowed to multiply delta y over delta x by delta u over delta u if we do not know that delta u is nonzero?
• In the second video at , shouldn't it be:
[lim x->c (u(x))] - u(c) ?
• Actually, with limit rules if you had lim(x-3) you could turn it into lim(x)-3, so Sal is using that property in reverse. He also says around that you can treat u(c) as a constant. Though I think it works even if you do it the way you explain.
• In the second video, we suppose that function is differentiable when rewriting the limit? Because continuity does not imply differentiability.
• There is no need to worry about differentiability in the second video because Sal never differentiates anything during the proof.
• It seems that using the same "reasoning" ("trick") used in the first video to demonstrate that differentiability implies continuity, we can prove statements which aren't (generally) true. Consider the following application to f(x) and an arbitrary number "a":

limit[x->c](f(x)-a)=
limit[x->c]((x-c)(f(x)-a)/(x-c))=
limit[x->c](x-c)limit[x->c](f(x)-a)=
0limit[x->c](f(x)-a)=
0

Is this a proof that limit of f(x) at an arbitrary "c" equals any arbitrary "a"?
(1 vote)
• The third step is wrong.
lim(𝑥 → 𝑐) [(𝑓(𝑥) − 𝑎)∕(𝑥 − 𝑐)] = ±∞ ∙ lim(𝑥 → 𝑐) [𝑓(𝑥) − 𝑎]
• Video 2 is supposed to prove that as the change in x APPROACHES 0, the change in u APPROACHES 0.

But at the end, Sal writes that the limit of the change in u is EQUAL to 0 as the change in x APPROACHES 0.

EQUAL and APPROACHES are not the same thing, are they? How can we conclude EQUALS from APPROACHING?
(1 vote)
• The sentences 'The limit as x goes c of Δu is 0' means the same thing as 'Δu approaches 0 as x approaches c'.

The quantity u is approaching 0, because the limit of u equals 0.