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## AP®︎ Calculus AB (2017 edition)

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

Lesson 9: Chain rule

# Proving the chain rule

Proving the chain rule for derivatives.
The chain rule tells us how to find the derivative of a composite function:
start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, close bracket, equals, f, prime, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, g, prime, left parenthesis, x, right parenthesis
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 $u$u is continuous at $x$x, then $\Delta u\to 0$delta, u, \to, 0 as $\Delta x\to 0$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!

\begin{aligned} \dfrac{d}{dx}\left[\dfrac{f(x)}{g(x)}\right]&=\dfrac{\dfrac{d}{dx}[f(x)]\cdot g(x)-f(x)\cdot\dfrac{d}{dx}[g(x)]}{[g(x)]^2} \\\\\\ &=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2} \end{aligned}