Unit 6: Differentiability: Mean Value Theorems, Taylor's Theorem & L'Hôpital's Rule

1. The Classical Definition of the Derivative

Definition 6.1 (Derivative at a Point): Let $I \subseteq \mathbb{R}$ be an open interval, $f: I \to \mathbb{R}$, and $c \in I$. The function $f$ is differentiable at $c$ if the limit of the difference quotient exists:

The real number $f'(c)$ is the derivative of $f$ at $c$.

2. Carathéodory's Equivalent Formulation

Constantin Carathéodory introduced a powerful reformulation that eliminates division by $x - c$, vastly simplifying proofs of the Chain Rule and multivariable extensions.

Theorem 6.1 (Carathéodory's Theorem): Let $f: I \to \mathbb{R}$ and $c \in I$. Then $f$ is differentiable at $c$ if and only if there exists a function $\phi: I \to \mathbb{R}$ that is continuous at $c$ such that:

In this case, $\phi(c) = f'(c)$.

Proof:

$(\implies)$ If $f'(c)$ exists, define:

Then $\lim_{x \to c} \phi(x) = f'(c) = \phi(c)$, so $\phi$ is continuous at $c$, and $f(x) - f(c) = \phi(x)(x - c)$ holds for all $x$. $(\impliedby)$ If such a continuous $\phi$ exists, then for $x \ne c$:

Taking limits as $x \to c$: $\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{x \to c} \phi(x) = \phi(c)$, so $f'(c) = \phi(c)$ exists. $\blacksquare$

3. Differentiability Implies Continuity

Theorem 6.2 (Differentiability Implies Continuity): If $f$ is differentiable at $c$, then $f$ is continuous at $c$.

Proof:

Using Carathéodory's formulation:

The Chain Rule via Carathéodory: Let $g$ be differentiable at $c$ and $f$ differentiable at $g(c)$. By Carathéodory: $g(x) - g(c) = \psi(x)(x - c)$ with $\psi(c) = g'(c)$, and $f(y) - f(g(c)) = \phi(y)(y - g(c))$ with $\phi(g(c)) = f'(g(c))$. Substituting $y = g(x)$:

Since composite continuous functions are continuous, $\lim_{x \to c} [\phi(g(x)) \psi(x)] = f'(g(c)) g'(c)$. Thus $(f \circ g)'(c) = f'(g(c)) g'(c)$ with zero division-by-zero complications! $\blacksquare$

1. Interior Local Extrema: Fermat's Theorem

Definition 6.2 (Local Extrema): Let $f: E \to \mathbb{R}$. A point $c \in E$ is a local maximum (resp. local minimum) if there exists $\delta > 0$ such that $f(x) \le f(c)$ (resp. $f(x) \ge f(c)$) for all $x \in (c - \delta, c + \delta) \cap E$.

Theorem 6.3 (Fermat's Interior Extremum Lemma): Let $f: (a, b) \to \mathbb{R}$ have a local extremum at an interior point $c \in (a, b)$. If $f$ is differentiable at $c$, then:

Line-by-Line Proof:

Assume without loss of generality that $c$ is a local maximum (the local minimum case is analogous). Then there exists $\delta > 0$ such that $f(x) \le f(c)$ for all $x \in (c - \delta, c + \delta) \subseteq (a, b)$.

1. Left-hand difference quotient ($x < c$):

For any $x \in (c - \delta, c)$, $x - c < 0$ and $f(x) - f(c) \le 0$. Therefore:

2. Right-hand difference quotient ($x > c$):

For any $x \in (c, c + \delta)$, $x - c > 0$ and $f(x) - f(c) \le 0$. Therefore:

Since $f$ is differentiable at $c$, both one-sided limits exist and equal $f'(c)$:

2. Darboux's Intermediate Value Theorem for Derivatives

Does the derivative of a function have to be continuous? No! For instance, $f(x) = x^2 \sin(1/x)$ for $x \ne 0$ with $f(0) = 0$ is everywhere differentiable, but its derivative oscillates wildly and is discontinuous at $x = 0$. However, Jean Gaston Darboux proved that every derivative still satisfies the Intermediate Value Property, even if it is completely discontinuous!

Theorem 6.4 (Darboux's Theorem): Let $f: [a, b] \to \mathbb{R}$ be differentiable on $[a, b]$. If $k$ is any real number strictly between $f'(a)$ and $f'(b)$, then there exists at least one $c \in (a, b)$ such that:

Proof:

Assume without loss of generality that $f'(a) < k < f'(b)$. Define the auxiliary function:

Since $f$ is differentiable on $[a, b]$, $g$ is continuous on $[a, b]$. By the Weierstrass Extreme Value Theorem (Theorem 5.4), $g$ attains its global minimum on $[a, b]$ at some point $c \in [a, b]$. We show that $c$ cannot be $a$ or $b$:

• At $a$: $g'(a) = f'(a) - k < 0$.

Since $g'(a) < 0$, $\lim_{x \to a^+} \frac{g(x) - g(a)}{x - a} < 0$, which means $g(x) < g(a)$ for all $x$ sufficiently close to $a$ on the right. Thus $a$ is not the minimum: $c \ne a$.

• At $b$: $g'(b) = f'(b) - k > 0$.

Since $g'(b) > 0$, $\lim_{x \to b^-} \frac{g(x) - g(b)}{x - b} > 0$, which means $g(x) < g(b)$ for all $x$ sufficiently close to $b$ on the left. Thus $b$ is not the minimum: $c \ne b$. Therefore, the minimum must occur at an interior point $c \in (a, b)$. By Fermat's Lemma (Theorem 6.3):

1. Rolle's Theorem

Theorem 6.5 (Rolle's Theorem): Let $f: [a, b] \to \mathbb{R}$ satisfy:

1. $f$ is continuous on the closed interval $[a, b]$.
2. $f$ is differentiable on the open interval $(a, b)$.
3. $f(a) = f(b)$.

Then there exists at least one point $c \in (a, b)$ such that $f'(c) = 0$.

Proof:

By the Extreme Value Theorem, $f$ attains both a maximum $M$ and a minimum $m$ on $[a, b]$.

• If $M = m$, $f$ is constant on $[a, b]$, so $f'(c) = 0$ for all $c \in (a, b)$.
• If $M \ne m$, since $f(a) = f(b)$, at least one of the extrema (either $M$ or $m$) must occur at an interior point $c \in (a, b)$.

By Fermat's Lemma, at this interior extremum, $f'(c) = 0$. $\blacksquare$

2. Lagrange's Mean Value Theorem

Theorem 6.6 (Lagrange's Mean Value Theorem - MVT): Let $f: [a, b] \to \mathbb{R}$ be continuous on $[a, b]$ and differentiable on $(a, b)$. Then there exists at least one point $c \in (a, b)$ such that:

Proof:

The secant line connecting $(a, f(a))$ and $(b, f(b))$ has equation:

Define the vertical distance function $h: [a, b] \to \mathbb{R}$:

1. $h$ is continuous on $[a, b]$ (linear combination of continuous functions).
2. $h$ is differentiable on $(a, b)$ with $h'(x) = f'(x) - \frac{f(b) - f(a)}{b - a}$.
3. $h(a) = f(a) - f(a) = 0$, and $h(b) = f(b) - f(b) = 0$.

Thus $h(a) = h(b) = 0$. By Rolle's Theorem, there exists $c \in (a, b)$ such that $h'(c) = 0$:

3. Cauchy's Generalized Mean Value Theorem

Theorem 6.7 (Cauchy's Mean Value Theorem): Let $f, g: [a, b] \to \mathbb{R}$ be continuous on $[a, b]$ and differentiable on $(a, b)$. Then there exists $c \in (a, b)$ such that:

If $g'(x) \ne 0$ on $(a, b)$, this is equivalently written as:

Proof:

Define the auxiliary function:

Notice that:

• $h(a) = f(b)g(a) - f(a)g(a) - g(b)f(a) + g(a)f(a) = f(b)g(a) - g(b)f(a)$.
• $h(b) = f(b)g(b) - f(a)g(b) - g(b)f(b) + g(a)f(b) = f(b)g(a) - g(b)f(a) = h(a)$.

$h$ satisfies Rolle's Theorem on $[a, b]$. Thus there exists $c \in (a, b)$ with $h'(c) = 0$:

1. The $0/0$ Indeterminate Form

Theorem 6.8 (L'Hôpital's Rule for $0/0$): Let $f$ and $g$ be differentiable on $(a, b) \setminus \{c\}$, where $c \in [a, b]$. Suppose:

1. $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$.
2. $g'(x) \ne 0$ for all $x \in (a, b) \setminus \{c\}$.
3. $\lim_{x \to c} \frac{f'(x)}{g'(x)} = L$ (where $L \in \mathbb{R}$ or $\pm \infty$).

Then:

Complete Line-by-Line Proof:

We extend $f$ and $g$ continuously to $c$ by defining $f(c) = 0$ and $g(c) = 0$. Since $\lim_{x \to c} f(x) = 0 = f(c)$ and $\lim_{x \to c} g(x) = 0 = g(c)$, both $f$ and $g$ are continuous at $c$. Let $x \in (c, b)$. Both $f$ and $g$ are continuous on $[c, x]$ and differentiable on $(c, x)$. By Cauchy's Generalized Mean Value Theorem (Theorem 6.7), there exists a point $\xi_x \in (c, x)$ such that:

Since $f(c) = 0$ and $g(c) = 0$:

As $x \to c^+$, the point $\xi_x \in (c, x)$ is squeezed towards $c$: $c < \xi_x < x \implies \lim_{x \to c^+} \xi_x = c$. Therefore:

An identical argument on $(a, c)$ establishes the left-hand limit $\lim_{x \to c^-} \frac{f(x)}{g(x)} = L$. Combining both one-sided limits proves the theorem. $\blacksquare$

2. The $\infty/\infty$ Indeterminate Form

Theorem 6.9 (L'Hôpital's Rule for $\infty/\infty$): If $\lim_{x \to c} |g(x)| = \infty$ and $\lim_{x \to c} \frac{f'(x)}{g'(x)} = L$, then:

Notice that we do not even need to assume that $\lim f(x) = \infty$!

1. Taylor Polynomials and Approximations

Definition 6.3 (Taylor Polynomial): Let $f$ be $n$-times differentiable at $x_0$. The $n$-th order Taylor polynomial of $f$ centered at $x_0$ is:

The remainder (error) is $R_n(x) = f(x) - P_n(x)$.

2. Taylor's Theorem with Lagrange Remainder

Theorem 6.10 (Taylor's Theorem with Lagrange Remainder): Let $f$ be $(n+1)$-times differentiable on an open interval $I$ containing $x_0$ and $x$. Then there exists some point $c$ strictly between $x_0$ and $x$ such that:

Complete Rigorous Proof:

Fix $x \ne x_0$. Define the constant $M$ such that:

Consider the auxiliary function $g: I \to \mathbb{R}$ defined for $t$ between $x_0$ and $x$:

Notice that:

• At $t = x$: $g(x) = f(x) - f(x) - 0 = 0$.
• At $t = x_0$: $g(x_0) = f(x) - P_n(x) - M(x - x_0)^{n+1} = 0$ (by definition of $M$).

Thus $g(t)$ is continuous on the closed interval between $x_0$ and $x$, and differentiable in between. Now we compute the derivative $g'(t)$ with respect to $t$. Differentiating the summation using the product rule:

Differentiating the $M$ term: $\frac{d}{dt}[-M(x - t)^{n+1}] = (n+1) M (x - t)^n$. Therefore:

By Rolle's Theorem (Theorem 6.5), there exists $c$ strictly between $x_0$ and $x$ such that $g'(c) = 0$:

Since $c \ne x$, $(x - c)^n \ne 0$. Dividing by $(x - c)^n$:

Substituting $M$ back into $R_n(x) = M(x - x_0)^{n+1}$ yields:

3. Cauchy and Integral Forms of the Remainder

• Cauchy Remainder Form:

• Integral Remainder Form (via integration by parts):