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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!

$\begin{array}{rl}\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]& =\frac{\frac{d}{dx}\left[f\left(x\right)\right]\cdot g\left(x\right)-f\left(x\right)\cdot \frac{d}{dx}\left[g\left(x\right)\right]}{\left[g\left(x\right){\right]}^{2}}\\ \\ \\ & =\frac{{f}^{\prime }\left(x\right)g\left(x\right)-f\left(x\right){g}^{\prime }\left(x\right)}{\left[g\left(x\right){\right]}^{2}}\end{array}$