Chain Rule

Let $f : I \to R$ and $g : J \to R$ where $I, J$ are intervals and $f (I) \subseteq J$ . Suppose $f$ is differentiable at $c$ and $g$ is differentiable at $f (c)$ . Then $(g \circ f)$ is differentiable at $c$ and is:

(g \circ f)^{'} (c) = g^{'} (f (c)) f^{'} (c)

It is tempting to use, in a similar way of deriving the Algebraic Differentiability Theorem, via the difference quotient:

\begin{aligned} \frac{g (f (x)) - g (f (c))}{x - c} & = \frac{g (f (x)) - g (f (c))}{f (x) - f (c)} \cdot \frac{f (x) - f (c)}{x - c} \end{aligned}

Which we cannot use directly. We'll have to do some work to justify the $f (x), f (c)$ bits in order to use this, which we do via $μ$ below.

Proof

For $y \in J$ define the function:

μ (y) = {\begin{cases} \frac{g (y) - g (f (c))}{y - f (c)} & y \neq f (c) \\ g^{'} (f (c)) & y = f (c) \end{cases}

Now notice that $μ$ is continuous at $f (c)$ ( $J$ is on an interval, and since $f (c)$ then:

lim_{y \to f (c)} μ (y) = lim_{y \to f (c)} \frac{g (y) - g (f (c))}{y - f (c)} = g^{'} (f (c)) = μ (f (c))

because $g$ itself is continuous).

Now also notice that $\forall y \in J$ , we claim that:

g (y) - g (f (c)) = μ (y) (y - f (c))

which is true since when $y \neq f (c)$ it's definitely true (multiply it out) and for $y = f (c)$ then the LHS becomes 0, which equals the RHS.

Now let $x \in I$ such that $x \neq c$ . Apply the above with $y = f (x)$ . Then:

\begin{aligned} g (f (x)) - g (f (c)) & = μ (f (x)) (f (x) - f (c)) \\ \frac{g (f (x)) - g (f (c))}{x - c} & = μ (f (x)) \cdot \frac{f (x) - f (c)}{x - c} \\ lim_{x \to c} \frac{g (f (x)) - g (f (c))}{x - c} & = lim_{x \to c} (μ (f (x)) \cdot \frac{f (x) - f (c)}{x - c}) \\ = lim_{x \to c} μ (f (x)) \cdot f^{'} (c) \end{aligned}

(note we used the Algebraic Limit Theorem for Functional Limits on the last step). Now because $f$ is continuous at $c$ and $μ$ is continuous, then $f \circ μ$ is continuous at $c$ and thus it's limit is $g (f (c))$ . As such then:

\begin{aligned} = μ (f (c)) \cdot f^{'} (c) \\ = g^{'} (f (c)) \cdot f^{'} (c) \end{aligned}

as desired.

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