Lecture 29 - Chain Rule

Announcements

Keep in mind:

Monday: No Class
Tuesday: Review
Wednesday: Exam III

Now we have enough info to derive the Chain Rule:

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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The chain rule helps especially to show when a function is differentiable. For example, can you find a function that's differentiable on $R$ who's derivative is not continuous on $R$ ? For example, the absolute value function $| x |$ does this kind-of at $0$ , but it isn't even differentiable at that point.

In fact, maybe it's a reasonable guess that it's impossible! To investigate it further let's look at some important examples.

Important Examples of Peculiarly Derived Functions

Consider the example 5 from before:

f_{0} (x) = {\begin{cases} \sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{cases}

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f(x) = sin(1/x)

It's not continuous at $x = 0$ but is continuous everywhere else. But by the Chain Rule then $f_{0} (x)$ it is differentiable at all $x \neq 0$ .

Now consider trying to make the function continuous, essentially squeezing the function towards 0 as $x \to 0$ . The idea is that:

f_{1} (x) = {\begin{cases} x \sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{cases}

If we plot this instead:

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f(x)= x*sin(1/x)

We can show that $f_{1} (x)$ is continuous at $x = 0$ because:

| f_{1} (x) - f_{1} (0) | = | x | | \sin (\frac{1}{x}) | \leq | x |

So if $0 < | x - 0 | = | x | < δ$ then choosing $δ < ε$ works:

| f_{1} (x) | \leq | x | < δ < ε

as desired.

Now by the Chain Rule then $f_{1}$ is differentiable at $x \neq 0$ , but what about $x = 0$ ? Let's check:

\begin{aligned} lim_{x \to 0} \frac{f_{1} (x) - f_{1} (0)}{x - 0} & = lim_{x \to 0} \frac{x \sin (\frac{1}{x})}{x} \\ = lim_{x \to 0} \sin (\frac{1}{x}) \end{aligned}

does not exist. But now we can try to go one step further by making $f_{1}$ more "flat" as $x \to 0$ . Define it as:

f_{2} (x) = {\begin{cases} x^{2} \sin (\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \end{cases}

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f(x) = x^2 * sin(1/x)

Will this smoothing work? Certainly it's continuous at $x = 0$ (just use the Algebraic Continuity Theorem here using $f_{1}$ ). For differentiability by the Chain Rule then $f_{2}$ is differentiable at all $x \neq 0$ . Notice that at this point:

f^{'} (x) = 2 x \sin (\frac{1}{x}) + x^{2} \cos (\frac{1}{x}) \cdot (\frac{- 1}{x^{2}}) = 2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x})

Now notice that as $x \to 0$ then the limit for $f^{'} (x)$ as $x \to 0$ doesn't exist. But this argument actually doesn't work for finding the derivative at 0! Directly using the definition of a derivative then we can show it actually is differentiable there. At 0:

\begin{aligned} f^{'} (0) & = lim_{x \to 0} \frac{f_{2} (x) - f_{2} (0)}{x - 0} \\ = lim_{x \to 0} \frac{x^{2} \sin (\frac{1}{x})}{x} \\ = lim_{x \to 0} x \sin (\frac{1}{x}) \\ = 0 \end{aligned}

Due to $f_{1} (x)$ being continuous at $0$ (giving the result of the limit) thus $f_{2}$ is differentiable over all $R$ .

Back to More Theorems

We can prove the Interior Extremum Theorem, also known as a Fermat's Theorem.

This is also known as Fermat's Theorem.

Interior Extremum Theorem

Let $f : (a, b) \to R$ . Suppose $f$ has a maximum or minimum at some $c \in (a, b)$ , and suppose $f$ is differentiable at $c$ . Then $f^{'} (c) = 0$ .

Proof

Without loss of generality, suppose $f (c)$ is a maximum. Then $\forall x \in (a, b)$ then $f (x) \leq f (c)$ . Now $a < c < b$ , so then we can make a sequence on the right and left of $c$ . Namely, since $f$ is differentiable at $c$ , then:
$\begin{aligned} f^{'} (c) & = lim_{x \to c} \frac{f (x) - f (c)}{x - c} \\ = lim_{x \to c^{-}} \frac{f (x) - f (c)}{x - c} \\ \geq 0 \end{aligned}$
Because $f (x) - f (x) \geq 0$ (using $f (c)$ is a maximum) and via the Limits and Order (Order Limit Theorem) (extending it to limits of functions). Similarly:
$\begin{aligned} f^{'} (c) & = lim_{x \to c} \frac{f (x) - f (c)}{x - c} \\ = lim_{x \to c^{+}} \frac{f (x) - f (c)}{x - c} \\ \leq 0 \end{aligned}$
As a result then $0 \leq f^{'} (c) \leq 0$ thus $f^{'} (c) = 0$ .

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