Chapter 05 The Derivative

5.2 Derivatives and the Intermediate Value Property

Definition 5.2.1 (Differentiability). Let $g : A → \mathbf{R}$ be a function defined on an interval $A$. Given $c ∈ A$, the derivative of $g$ at $c$ is defined by

$$g'(c) = \lim_{x \to c} \frac{ g(x) − g(c) }{ x−c }$$

Provided this limit exists. In this case we say $g$ is differentiable at $c$. If $g$ exists for all points $c ∈ A$, we say that $g$ is differentiable on $A$.

Theorem 5.2.3. If $g : A → \mathbf{R}$ is differentiable at a point $c ∈ A$, then $g$ is continuous at $c$ as well.

Proof: Assume $g'(c) = d$, that means given $\epsilon$, we can find $\delta$, if $x \in U_{\delta}(c)$, then $\left| \frac{g(x) - g(c)}{x - c} \right| < \epsilon$. let $\delta' = \min \{1, \delta\}$. Then if $x \in U_{\delta'}(c)$,

$$\left| g(x) - g(c) \right| < \left| \frac{g(x) - g(c)}{x - c} \right| < \epsilon$$

So $g$ is continuous at $c$ as well. $\square$

Theorem 5.2.5 (Chain Rule). Let $f : A → R$ and $g : B → R$ satisfy $f(A) ⊆ B$ so that the composition $g ◦ f$ is defined. If $f$ is differentiable at $c∈A$ and if $g$ is differentiable at $f(c)∈B$,then $g◦f$ is differentiable at $c$ with $(g ◦ f)'(c) = g'(f(c)) · f'(c)$.

Proof:

$$(g ◦ f)'(c) = \lim_{x \to c} \frac{(g ◦ f)(x)- (g ◦ f)(c)}{x - c} \\ = \lim_{x \to c} \frac{g(f(x))- g(f(c))}{f(x) - f(c)} \cdot \frac{f(x) - f(c)}{x - c} \\ $$

Let

$$d(y) = \frac{g(y) - g(f(c))}{y - f(c)}$$

Define $d(y) = g'(f(c))$. So $d$ is continuous at $f(c)$. So we have $g(y) - g(f(c)) = d(y)(y - c)$, $\forall y \in B$.

So we have $g(f(t)) - g(f(c)) = d(f(t))(f(t) - c)$, $\forall t \in A$

So

$$\frac{g(f(t))- g(f(c))}{t - c}\\ = \frac{d(f(t))(f(t) - c)}{t - c}\\ = d(f(t))\frac{(f(t) - c)}{t - c}$$

Because $f(t)$ is continuous at $c$, $d(y)$ is continuous at $f(c)$, so $d(f(t))$ is continuous at $t$ based on Theorem 4.3.9. $\frac{(f(t) - c)}{t - c}$ is also continuous at $t$. With Corollary 4.2.4 (Algebraic Limit Theorem for Functional Limits).

$$\lim_{t \to c} d(f(t))\frac{(f(t) - c)}{t - c} \\ = \lim_{t \to c}d(f(t)) \cdot \lim_{t \to c} \frac{(f(t) - c)}{t - c}\\ = g'(f(c)) \cdot f'(c)$$ $\square$

Darboux’s Theorem

Theorem 5.2.6 (Interior Extremum Theorem). Let $f$ be differentiable on an open interval $(a, b)$. If $f$ attains a maximum value at some point $c ∈ (a, b)$ (i.e., $f(c) ≥ f(x)$ for all $x ∈ (a,b)$), then $f'(c) = 0$. The same is true if $f(c)$ is a minimum value.

Proof:

$$\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{x \to c^+} \frac{f(x) - f(c)}{x - c} \le 0 \\ \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{x \to c^-} \frac{f(x) - f(c)}{x - c} \ge 0 \\ $$

So $f'(c) = 0$. $\square$

Theorem 5.2.7 (Darboux's Theorem). If $f$ is differentiable on an interval $[a,b]$, and if $\alpha$ satisfies $f'(a) < \alpha < f'(b)$ (or $f'(a) > \alpha > f'(b)$), then there exists a point $c ∈ (a, b)$ where $f'(c) = \alpha$.

Proof: Consider $g(x) = f(x) - \alpha x$.

$$g'(a) = f'(a) - \alpha < 0 \text{ and } g'(b) = f'(b) - \alpha > 0$$

• If $g(a) >= g(b)$, since $g'(b) > 0$, then exists $\delta$, such that if $x \in (b-\delta , b)$, $g(x) < g(b)$. Then $g$ attains a minimum value at some point $c ∈ (a, b)$. So $g'(c) = 0$ and $f'(c) = \alpha$.
• If $g(a) < g(b)$, since $g'(a) < 0$, then exists $\delta$, such that if $x \in (a, a+\delta)$, $g(x) < g(a)$. Then $g$ attains a minimum value at some point $c ∈ (a, b)$. So $g'(c) = 0$ and $f'(c) = \alpha$.

$\square$

5.3 The Mean Value Theorems

• The ease of the proof, however, is misleading, as it is built on top of some hard-fought accomplishments from the study of limits and continuity.
• The Mean Value Theorem is the cornerstone of the proof for almost every major theorem pertaining to differentiation.
  • We will use it to prove L’Hospital’s rules regarding limits of quotients of differentiable functions.
  • A rigorous analysis of how infinite series of functions behave when differentiated requires the Mean Value Theorem (Theorem 6.4.3)
  • It is the crucial step in the proof of the Fundamental Theorem of Calculus (Theorem 7.5.1)
  • It is also the fundamental concept underlying Lagrange’s Remainder Theorem (Theorem 6.6.3) which approximates the error between a Taylor polynomial and the function that generates it.

Theorem 5.3.1 (Rolle’s Theorem).

Let $f : [a, b] → \mathbf{R}$ be continuous on $[a, b]$ and differentiable on $(a,b)$. If $f(a) = f(b)$, then there exists a point $c ∈ (a,b)$ where $f'(c) = 0$.

Sketch of Proof: $f$ is a continuous, so it attains its maximum or minimum value at some point $c ∈ (a,b)$. Then $f'(c) = 0$ from the Interior Extremum Theorem (Theorem 5.2.6).

Theorem 5.3.2 (Mean Value Theorem).

If $f : [a, b] → \mathbf{R}$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists a point $c ∈ (a, b)$ where

$$f'(c)= \frac{f(b)-f(a)}{b-a}$$

Sketch of Proof: Let

$$g(x) = \frac{f(b)-f(a)}{b-a} (x-a) - f(x)$$

Then $g(a) = -f(a)$, $g(b) = -f(a)$. Then we can apply Rolle theorem:

$$g'(c) = \frac{f(b)-f(a)}{b-a} - f'(c) = 0$$

So

$$f'(c)= \frac{f(b)-f(a)}{b-a}$$

Theorem 5.3.5 (Generalized Mean Value Theorem).

If $f$ and $g$ are continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists a point $c ∈ (a, b)$ where

$$[f (b) − f (a)]g' (c) = [g(b) − g(a)]f' (c)$$

If $g'$ is never zero on $(a, b)$, then the conclusion can be stated as

$$\frac{f(b) - f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}$$

Proof: See exercise 5.3.5.

$\square$

Theorem 5.3.6 (L’Hospital’s Rule: 0/0 case).

Let $f$ and $g$ be continuous on an interval containing $a$, and assume $f$ and $g$ are differentiable on this interval with the possible exception of the point $a$. If $f(a) = g(a) = 0$ and $g'(x) \not =0$ for all $x \not =a$, then

$$\lim_{x \to a} \frac{f'(x)}{g'(x)} = L \text{ implies } \lim_{x \to a} \frac{f(x)}{g(x)} = L$$

Proof: See Exercise 5.3.11.

Theorem 5.3.8 (L’Hospital’s Rule: $∞/∞$ case).

Assume $f$ and $g$ are differentiable on $(a, b)$ and that $g'(x) \not = 0$ for all $x ∈ (a, b)$. If $\lim_{x→a} g(x) = ∞$ (or $−∞$), then

$$\lim_{x \to a} \frac{f'(x)}{g'(x)} = L \text{ implies } \lim_{x \to a} \frac{f(x)}{g(x)} = L$$

Proof: We can find $\delta_1$ such that

$$\left| \frac{f'(x)}{g'(x)} - L \right| < \epsilon /2$$

Let $t = a + \delta_1$, for any $x \in (a, t)$, we have

$$L - \epsilon / 2 < \frac{ f(x) - f(t) }{g(x) - g(t)} < L + \epsilon / 2$$

We times $1 - \frac{g(t)}{g(x)}$ and get

$$(1 - \frac{g(t)}{g(x)})(L - \epsilon / 2) < (1 - \frac{g(t)}{g(x)})(\frac{ f(x) - f(t) }{g(x) - g(t)}) < (1 - \frac{g(t)}{g(x)})(L + \epsilon / 2) \\ L - \epsilon / 2 - \frac{ L g(t) - \epsilon / 2 g(t) - f(t) }{g(x)} < \frac{g(x)}{f(x)} < L + \epsilon / 2 + \frac{ -L g(t) - \epsilon / 2 g(t) + f(t) }{g(x)}$$

As long as $x$ is close to $a$ enough,

$$L - \epsilon < \frac{ f(x) }{g(x)} < L + \epsilon$$

So

$$\lim_{x \to a} \frac{f(x)}{g(x)} = L$$

$\square$