CommeUnJeu · L1 PCSI

Differentiability

⌚ ~131 min ▢ 16 blocks ✓ 52 exercises ➣ Prerequisites : Limits and continuity, Real functions: recap

Differentiability formalises the secondary-school slope of a tangent line as the limit of a difference quotient. From this single notion springs a whole calculus: algebraic rules, the chain rule, the derivative of an inverse map, then three pillars of analysis --- Fermat's theorem, Rolle's theorem, and the mean value theorem. These produce the mean value inequality, the link sign-of-$f'$ $\leftrightarrow$ monotonicity, and the theorem of the limit of the derivative for rigorous $C^1$ extensions. We close with iterated derivatives, the class $C^k$, the Leibniz formula for $(fg)^{(n)}$, and a brief extension to complex-valued functions.
Endpoint convention used throughout. When $a \in I$ is an interior point of $I$, « differentiable at $a$ » is the bilateral notion (limit from both sides). When $a$ is an endpoint of $I$, it is the one-sided notion (limit from inside $I$). Most global theorems use the hypothesis « $f$ continuous on $I$, differentiable on $\mathring{I}$ »: differentiability is then only required at interior points, while endpoint behavior is handled by continuity.
Forward references to Standard functions. This chapter uses polynomials, rational functions, $\sqrt{\cdot}$, $|\cdot|$, $\sqrt[3]{\cdot}$ as a self-contained example toolbox. Wherever an example invokes $\exp$, $\sin$, $\cos$, the derivative is admitted from Real functions: lycée recap; derivations of $\ln$, $\arcsin$, $\arctan$, hyperbolic inverses are deferred to Standard functions, where they will use P2.5 of this chapter.

I Derivative at a point --- definition and tangent

The derivative at $a$ measures the instantaneous rate of change of $f$ at $a$, generalising the secondary-school slope of a tangent line. Two equivalent forms: the limit of the difference quotient and the first-order asymptotic expansion ($DL_1$). Lateral derivatives are defined by restricting the increment $h$ to positive or negative values; at an interior point, differentiability means that both lateral limits exist and are equal.

Definition — Derivative at a point

Let $f : I \to \mathbb{R}$ and $a \in I$. For every $h \in \mathbb{R}^*$ such that $a + h \in I$, the difference quotient of $f$ at $a$ with increment $h$ is $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h}. $$ We say that $f$ is differentiable at $a$ if $\tau_a(h)$ admits a finite limit as $h \to 0$ with $a + h \in I$. This limit is called the derivative of $f$ at $a$ and is denoted $$ \textcolor{colordef}{f'(a)} \qquad \text{or} \qquad \textcolor{colordef}{\frac{df}{dx}(a)}. $$ Equivalently, by setting $x = a + h$, this amounts to requiring the finite limit of $$ \frac{f(x) - f(a)}{x - a} \quad \text{as } x \to a. $$

Proposition — Equivalent characterization via $DL_1$

$f$ is differentiable at $a$ if and only if there exist $\ell \in \mathbb{R}$ and a function $\varepsilon$ defined on $\{h \in \mathbb{R} : a + h \in I\}$ with $\varepsilon(h) \to 0$ as $h \to 0$, such that $$ \textcolor{colorprop}{f(a + h) = f(a) + \ell h + h \varepsilon(h)} \qquad \text{for every } h \text{ with } a + h \in I. $$ In this case $\ell = f'(a)$.

Proof

$(\Rightarrow)$ Suppose $\tau_a(h) \to \ell$ as $h \to 0$. For $h$ such that $a + h \in I$ and $h \ne 0$, set $$ \varepsilon(h) = \tau_a(h) - \ell = \frac{f(a + h) - f(a)}{h} - \ell. $$ Then $\varepsilon(h) \to 0$ as $h \to 0$ and $f(a + h) - f(a) = \ell h + h \varepsilon(h)$. We extend by $\varepsilon(0) = 0$, which does not affect the limit.
$(\Leftarrow)$ If $f(a + h) = f(a) + \ell h + h \varepsilon(h)$ with $\varepsilon(h) \to 0$, then for $h \ne 0$, $\tau_a(h) = \ell + \varepsilon(h) \to \ell$, so $f$ is differentiable at $a$ and $f'(a) = \ell$.

Definition — Tangent line

Suppose $f$ is differentiable at $a$. The tangent line to the graph of $f$ at $a$ is the line of equation $$ \textcolor{colordef}{y = f(a) + f'(a)(x - a)}. $$ If $a$ is an endpoint of $I$, we speak of a half-tangent line on the side that lies inside $I$. If $\tau_a(h)$ admits no finite limit but tends to $\pm \infty$ as $h \to 0$, we speak of a vertical tangent of equation $x = a$ (a vertical half-tangent at an endpoint).

Figure --- secant lines converging to the tangent

The secant lines from $(a, f(a))$ to $(a + h, f(a + h))$ have slope $\tau_a(h)$. As $h \to 0$, the slope converges to $f'(a)$ and the secants converge to the tangent line.

Example

Compute $f'(a)$ for $f(x) = x^2$, for every $a \in \mathbb{R}$.

Answer

For $h \ne 0$, $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h} = \frac{(a + h)^2 - a^2}{h} = \frac{2 a h + h^2}{h} = 2 a + h. $$ As $h \to 0$, $\tau_a(h) \to 2 a$. Hence $f'(a) = 2 a$.

Example

For $f : [0, +\infty[ \to \mathbb{R}$, $f(x) = \sqrt{x}$, study differentiability at $0$.

Answer

For $h > 0$, $$ \tau_0(h) = \frac{\sqrt{0 + h} - \sqrt{0}}{h} = \frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}} \to +\infty \quad (h \to 0^+). $$ The limit is infinite, so $f$ is not differentiable at $0$, but the graph admits a vertical half-tangent of equation $x = 0$ at the origin.

Definition — Left and right derivatives

Let $a \in I$. We say $f$ is right-differentiable at $a$ if $\tau_a(h)$ admits a finite limit as $h \to 0^+$ with $a + h \in I$. This limit is denoted $\textcolor{colordef}{f'_d(a)}$.
We say $f$ is left-differentiable at $a$ if $\tau_a(h)$ admits a finite limit as $h \to 0^-$ with $a + h \in I$. This limit is denoted $\textcolor{colordef}{f'_g(a)}$.
If $a$ is an interior point of $I$, then $f$ is differentiable at $a$ if and only if $f'_g(a)$ and $f'_d(a)$ exist and are equal; the common value is $f'(a)$. At an endpoint, only the derivative from inside $I$ is considered.

Example

Show that $f(x) = |x|$ is not differentiable at $0$.

Answer

For $h > 0$, $\tau_0(h) = |h|/h = 1$, so $f'_d(0) = 1$. For $h < 0$, $\tau_0(h) = |h|/h = -1$, so $f'_g(0) = -1$. Since $f'_g(0) \ne f'_d(0)$, $f$ is not differentiable at $0$.

Proposition — Differentiable implies continuous

If $f$ is differentiable at $a$, then $\textcolor{colorprop}{f}$ is continuous at $a$.

Proof

For $h \ne 0$ with $a + h \in I$, $$ f(a + h) - f(a) = h \, \tau_a(h). $$ As $h \to 0$, $h \to 0$ and $\tau_a(h) \to f'(a) \in \mathbb{R}$, so $h \, \tau_a(h) \to 0$ by the product rule for function limits (P3.1 of Limits and continuity). Hence $f(a + h) \to f(a)$, which is continuity at $a$.

Example

The converse of P1.2 is false: $f(x) = |x|$ is continuous at $0$ (Ex1.3 above) but not differentiable at $0$.

Definition — Differentiability on an interval

$f$ is differentiable on $I$ if $f$ is differentiable at every point of $I$ (bilateral at interior points, unilateral at endpoints, per the endpoint convention). The map $f' : I \to \mathbb{R}$, $a \mapsto f'(a)$, is called the derivative function of $f$.

Method — Establishing differentiability at a point

Three classical approaches:

Direct via the difference quotient. Compute $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h} $$ and check whether it has a finite limit as $h \to 0$.
Via $DL_1$. Find $\ell \in \mathbb{R}$ and $\varepsilon$ with $\varepsilon(h) \to 0$ such that $f(a+h) = f(a) + \ell h + h \varepsilon(h)$. Then $f'(a) = \ell$.
Via the theorem of the limit of the derivative (T5.2 below). If $f$ is continuous at $a$ and differentiable on $I \setminus \{a\}$, compute $\lim_a f'$.

Skills to practice

Computing a derivative from the definition
Studying lateral differentiability
Using the $DL_1$ characterization of differentiability

II Algebraic operations on derivatives

Linearity, product, quotient, chain rule, derivative of the inverse: the same toolbox as the lycée, but with rigorous proofs. The derivative of the inverse map (P2.5) is the genuinely new atom at this level; it underlies the rigorous derivatives of $\arcsin$, $\arctan$, $\ln$ to be proved in Standard functions.

Proposition — Linearity

Let $f, g : I \to \mathbb{R}$ be differentiable at $a \in I$, and $\lambda, \mu \in \mathbb{R}$. Then $\lambda f + \mu g$ is differentiable at $a$ and $$ \textcolor{colorprop}{(\lambda f + \mu g)'(a) = \lambda f'(a) + \mu g'(a)}. $$

Proof

For $h \ne 0$ with $a + h \in I$, $$ \tau_a^{\lambda f + \mu g}(h) = \lambda \, \tau_a^f(h) + \mu \, \tau_a^g(h). $$ Passing to the limit as $h \to 0$ gives, by linearity of the function-limit (Limits and continuity P3.1), $\tau_a^{\lambda f + \mu g}(h) \to \lambda f'(a) + \mu g'(a)$.

Proposition — Product rule

Let $f, g : I \to \mathbb{R}$ be differentiable at $a \in I$. Then $f g$ is differentiable at $a$ and $$ \textcolor{colorprop}{(f g)'(a) = f'(a) g(a) + f(a) g'(a)}. $$

Proof

For $h \ne 0$ with $a + h \in I$, $$ f(a + h) g(a + h) - f(a) g(a) = (f(a + h) - f(a)) g(a + h) + f(a) (g(a + h) - g(a)). $$ Dividing by $h$ gives $$ \tau_a^{f g}(h) = \tau_a^f(h) \cdot g(a + h) + f(a) \cdot \tau_a^g(h). $$ As $h \to 0$: $\tau_a^f(h) \to f'(a)$; $g(a + h) \to g(a)$ by continuity of $g$ at $a$ (P1.2); $\tau_a^g(h) \to g'(a)$. Hence $\tau_a^{f g}(h) \to f'(a) g(a) + f(a) g'(a)$.

Proposition — Quotient rule

Let $f, g : I \to \mathbb{R}$ be differentiable at $a \in I$, with $g(a) \ne 0$. Then $f/g$ is defined on a neighborhood of $a$ in $I$, differentiable at $a$, and $$ \textcolor{colorprop}{(f/g)'(a) = \frac{f'(a) g(a) - f(a) g'(a)}{g(a)^2}}. $$

Proof

First, the derivative of $1/g$ at $a$. By continuity of $g$ at $a$ and $g(a) \ne 0$, $g$ does not vanish on a neighborhood of $a$. For small $h$ with $a + h$ in this neighborhood and $h \ne 0$, $$ \frac{1}{g(a + h)} - \frac{1}{g(a)} = \frac{g(a) - g(a + h)}{g(a + h) \, g(a)}. $$ Dividing by $h$: $$ \tau_a^{1/g}(h) = -\frac{\tau_a^g(h)}{g(a + h) \, g(a)} \to -\frac{g'(a)}{g(a)^2}. $$ Hence $1/g$ is differentiable at $a$ with $(1/g)'(a) = -g'(a)/g(a)^2$. The case $f/g = f \cdot (1/g)$ follows from the product rule (P2.2).

Example

By induction on $n \in \mathbb{N}$, show that $(x^n)' = n x^{n - 1}$ on $\mathbb{R}$.

Answer

Case $n = 0$: $f(x) = x^0 = 1$ is constant, so $f'(x) = 0$. (The formula $(x^n)' = n x^{n-1}$ is interpreted at $n = 0$ as $0$; the expression $x^{-1}$ does not need to be defined.) Case $n \ge 1$, by induction. Base $n = 1$: $f(x) = x$, $\tau_a(h) = ((a + h) - a)/h = 1 \to 1$, so $f'(a) = 1 = 1 \cdot x^0$. Heredity: assume $(x^n)' = n x^{n-1}$ for some $n \ge 1$. Then by the product rule (P2.2), $(x^{n+1})' = (x \cdot x^n)' = 1 \cdot x^n + x \cdot n x^{n-1} = x^n + n x^n = (n+1) x^n$.

Example

Compute the derivative of $f(x) = (x^3 - 2 x + 1)/(x^2 + 1)$ on $\mathbb{R}$.

Answer

$x^2 + 1 > 0$ never vanishes. By the quotient rule: $$ f'(x) = \frac{(3 x^2 - 2)(x^2 + 1) - (x^3 - 2 x + 1)(2 x)}{(x^2 + 1)^2} = \frac{x^4 + 5 x^2 - 2 x - 2}{(x^2 + 1)^2}. $$

Proposition — Chain rule

Let $f : I \to \mathbb{R}$ with $f(I) \subset J$, and $g : J \to \mathbb{R}$. If $f$ is differentiable at $a \in I$ and $g$ is differentiable at $b = f(a) \in J$, then $g \circ f$ is differentiable at $a$ and $$ \textcolor{colorprop}{(g \circ f)'(a) = g'(f(a)) \cdot f'(a)}. $$

Proof

Define an auxiliary function $\tau : J \to \mathbb{R}$ by $$ \tau(y) = \begin{cases} (g(y) - g(b))/(y - b) & \text{if } y \ne b, \\ g'(b) & \text{if } y = b. \end{cases} $$ By the differentiability of $g$ at $b$, $\tau(y) \to g'(b) = \tau(b)$ as $y \to b$, so $\tau$ is continuous at $b$. By definition, $g(y) - g(b) = \tau(y) (y - b)$ holds for all $y \in J$ (including $y = b$, both sides being $0$).
Substitute $y = f(a + h)$ for $h$ such that $a + h \in I$: $$ g(f(a + h)) - g(f(a)) = \tau(f(a + h)) \, (f(a + h) - f(a)). $$ For $h \ne 0$, divide by $h$: $$ \tau_a^{g \circ f}(h) = \tau(f(a + h)) \cdot \tau_a^f(h). $$ As $h \to 0$: $f$ is continuous at $a$ (P1.2) so $f(a + h) \to b$, and $\tau$ is continuous at $b$ so $\tau(f(a + h)) \to g'(b)$; finally $\tau_a^f(h) \to f'(a)$. The product tends to $g'(b) f'(a)$.

Example

Compute the derivatives of $(x^3 + 1)^5$ and $\sqrt{x^2 + 1}$ on $\mathbb{R}$ via the chain rule.

Answer

For $f(x) = (x^3 + 1)^5$: take $u(x) = x^3 + 1$ ($u'(x) = 3 x^2$), $v(y) = y^5$ ($v'(y) = 5 y^4$). By chain rule, $f'(x) = v'(u(x)) \cdot u'(x) = 5 (x^3 + 1)^4 \cdot 3 x^2 = 15 x^2 (x^3 + 1)^4$.
For $g(x) = \sqrt{x^2 + 1}$: take $u(x) = x^2 + 1$ ($u'(x) = 2 x$), $v(y) = \sqrt{y}$ defined on $y > 0$. Derivative of $v$ at $a > 0$: for $h \ne 0$ with $a + h > 0$, $$ \frac{\sqrt{a + h} - \sqrt{a}}{h} = \frac{(\sqrt{a + h} - \sqrt{a})(\sqrt{a + h} + \sqrt{a})}{h (\sqrt{a + h} + \sqrt{a})} = \frac{1}{\sqrt{a + h} + \sqrt{a}}. $$ As $h \to 0$, this tends to $1/(2 \sqrt{a})$, so $v'(a) = 1/(2 \sqrt{a})$. Then $g'(x) = v'(u(x)) \cdot u'(x) = (2 x)/(2 \sqrt{x^2 + 1}) = x/\sqrt{x^2 + 1}$.

Proposition — Derivative of the inverse map

Let $I$ be an interval, $f : I \to \mathbb{R}$ continuous and strictly monotone on $I$, $J = f(I)$. Suppose $f$ is differentiable at $a \in I$ with $f'(a) \ne 0$. Then $f^{-1} : J \to I$ is differentiable at $b = f(a)$ and $$ \textcolor{colorprop}{(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{f'(f^{-1}(b))}}. $$

Proof

Step 1: $f^{-1}$ is continuous at $b$. By the bijection theorem (Limits and continuity T7.2), $f^{-1} : J \to I$ is continuous and strictly monotone on $J$.
Step 2: limit of the inverse difference quotient. Take $k$ such that $b + k \in J$ and $k \ne 0$, and set $h = f^{-1}(b + k) - a$, so $a + h = f^{-1}(b + k) \in I$ and $h \ne 0$ (strict monotonicity of $f^{-1}$ gives $h \ne 0$ when $k \ne 0$). Then $f(a + h) = b + k$, so $f(a + h) - f(a) = k$, hence $$ \tau_b^{f^{-1}}(k) = \frac{f^{-1}(b + k) - f^{-1}(b)}{k} = \frac{h}{f(a + h) - f(a)} = \frac{1}{\tau_a^f(h)}. $$ As $k \to 0$, $h = f^{-1}(b + k) - a \to 0$ by continuity of $f^{-1}$ at $b$ (Step 1); hence $\tau_a^f(h) \to f'(a) \ne 0$, and $1/\tau_a^f(h) \to 1/f'(a)$.
Step 3: conclusion. $f^{-1}$ is differentiable at $b$ and $(f^{-1})'(b) = 1/f'(a) = 1/f'(f^{-1}(b))$.

Figure --- symmetry between graphs of $f$ and $f^{-1}$

The graphs of $f$ and $f^{-1}$ are symmetric about the line $y = x$. The tangent of $f$ at $(a, b)$ and the tangent of $f^{-1}$ at $(b, a)$ are mirror images. When $f'(a) = 0$ (horizontal tangent of $f$), the tangent of $f^{-1}$ at $(b, a)$ is vertical: $f^{-1}$ is not differentiable at $b$ but admits a vertical tangent.

Example

Show that $g : \mathbb{R} \to \mathbb{R}$, $g(y) = \sqrt[3]{y} = y^{1/3}$, is differentiable on $\mathbb{R} \setminus \{0\}$ and compute $g'(b)$ for $b \ne 0$.

Answer

$g$ is the inverse of $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^3$ (continuous, strictly increasing on $\mathbb{R}$). $f$ is differentiable on $\mathbb{R}$ with $f'(x) = 3 x^2$, which vanishes only at $x = 0$. For $b \ne 0$, let $a = g(b) = \sqrt[3]{b} \ne 0$, then $f'(a) = 3 a^2 \ne 0$ and P2.5 gives $$ g'(b) = \frac{1}{f'(a)} = \frac{1}{3 a^2} = \frac{1}{3 (\sqrt[3]{b})^2}. $$ At $b = 0$: $f'(0) = 0$, the hypothesis of P2.5 fails; $g$ admits a vertical tangent at $0$ (the tangent of $f$ at $0$ being horizontal).
Forward reference. This same method, applied with $f = \exp$, $f = \sin_{|[-\pi/2, \pi/2]}$, $f = \tan_{|]-\pi/2, \pi/2[}$, will give rigorously $(\ln)' = 1/x$, $(\arcsin)' = 1/\sqrt{1 - x^2}$, $(\arctan)' = 1/(1 + x^2)$ in Standard functions.

Method — Compute a derivative using the operation toolbox

For a function $f$ built from elementary blocks:

Identify the building blocks (polynomial, rational, $\sqrt{\cdot}$, composition, …).
Apply linearity / product / quotient / chain / inverse rules in the order suggested by the structure.
Simplify the result and state the domain of validity (where denominators do not vanish, etc.).

Skills to practice

Computing derivatives with the operation toolbox
Differentiating the inverse map

III Local extrema and Fermat's theorem

At an interior point of $I$ where $f$ is differentiable, an extremum forces $f'$ to vanish: this is Fermat's theorem, the bridge from differentiability to the next section on Rolle's theorem and the mean value theorem. The secondary-school notion of « stationary point » becomes here the « critical point ». Crucial: the hypothesis « interior point » is not removable, and the converse is false.

Definition — Local extremum

$f : I \to \mathbb{R}$, $a \in I$. We say $a$ is a local maximum (resp. local minimum) of $f$ if there exists $\eta > 0$ such that for all $x \in I \cap [a - \eta, a + \eta]$, $f(x) \le f(a)$ (resp. $\ge$). A local extremum is a local maximum or local minimum.

Definition — Critical point

$f : I \to \mathbb{R}$ differentiable at $a \in I$. We say $a$ is a critical point of $f$ if $\textcolor{colordef}{f'(a) = 0}$.

Theorem — Fermat

Let $f : I \to \mathbb{R}$ and $a$ an interior point of $I$ where $f$ is differentiable. If $a$ is a local extremum of $f$, then $\textcolor{colorprop}{f'(a) = 0}$.

Proof

Treat the case of a local maximum; the local minimum case is symmetric (replace $f$ by $-f$). Let $\eta > 0$ such that $[a - \eta, a + \eta] \subset I$ (possible because $a$ is interior) and $f(a + h) \le f(a)$ for every $h \in [-\eta, \eta]$.
For $0 < h \le \eta$, we have $f(a + h) \le f(a)$, so $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h} \le 0. $$ Passing to the limit as $h \to 0^+$ (passage to the limit of a large inequality for function limits, Limits and continuity P4.1) gives $f'_d(a) \le 0$.
For $-\eta \le h < 0$, we again have $f(a + h) \le f(a)$, but now $h < 0$, so $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h} \ge 0. $$ Passing to the limit as $h \to 0^-$ gives $f'_g(a) \ge 0$.
Since $f$ is differentiable at the interior point $a$, the two lateral derivatives are equal to $f'(a)$. Hence $f'(a) \le 0$ and $f'(a) \ge 0$, so $f'(a) = 0$.

Example

Verify Fermat on $f(x) = x^2$ at $a = 0$.

Answer

$f$ admits a global (hence local) minimum at $a = 0$ (since $f(x) \ge 0 = f(0)$). $a = 0$ is an interior point of $I = \mathbb{R}$. $f$ is differentiable at $0$ with $f'(0) = 0$ (Ex1.1). Consistency with Fermat.

Example

Counter-example 1 --- converse of Fermat is false. Show that $f(x) = x^3$ satisfies $f'(0) = 0$ but $0$ is not a local extremum of $f$.

Answer

$f'(x) = 3 x^2$ so $f'(0) = 0$. But for any $\eta > 0$, $f(-\eta/2) = -\eta^3/8 < 0 < \eta^3/8 = f(\eta/2)$, and $f(0) = 0$, so $0$ is neither a local maximum nor a local minimum of $f$. The point $0$ is a critical point of $f$ but not an extremum (it is an inflection point with horizontal tangent).

Example

Counter-example 2 --- the « interior » hypothesis is essential. Consider $f : [0, 1] \to \mathbb{R}$, $f(x) = x$. The maximum of $f$ is reached at $1$, an endpoint of $[0, 1]$ (not an interior point), so Fermat does not apply. In fact, for $h < 0$ with $1 + h \in [0, 1]$, $$ \tau_1(h) = \frac{f(1 + h) - f(1)}{h} = \frac{(1 + h) - 1}{h} = 1. $$ Thus $f'_g(1) = 1 \ne 0$.

Skills to practice

Locating extrema via critical points

IV Rolle's theorem and the mean value theorem

From Fermat we deduce Rolle, then the mean value theorem. Each is a global consequence of the extreme value theorem + Fermat. The mean value theorem is the workhorse: most subsequent inequalities (mean value inequality, monotonicity, limit-of-derivative) follow from it.

Theorem — Rolle

Let $a < b$ and $f : [a, b] \to \mathbb{R}$ continuous on $[a, b]$, differentiable on the open interval $]a, b[$, with $f(a) = f(b)$. Then there exists $c \in {]}a, b{[}$ such that $\textcolor{colorprop}{f'(c) = 0}$.

Proof

By the extreme value theorem (Limits and continuity T7.1), $f$ admits a maximum $M$ and a minimum $m$ on the segment $[a, b]$. Distinguish two cases.

Case 1: $M = m$. Then $f$ is constant on $[a, b]$, hence $f' \equiv 0$ on $]a, b[$, and any $c \in {]}a, b{[}$ works.
Case 2: $M \ne m$. Then $M > m$. If both extremes were reached only at the endpoints $a$ and $b$, both would equal $f(a) = f(b)$, contradicting $M \ne m$. Hence at least one of $M, m$ is reached at an interior point $c \in {]}a, b{[}$. Then $c$ is a local extremum (in fact a global one), $c$ is interior, and $f$ is differentiable at $c$, so by Fermat T3.1, $f'(c) = 0$.

Figure --- Rolle

With $f(a) = f(b)$, the chord is horizontal. Rolle guarantees at least one interior point $c$ where the tangent is also horizontal.

Example

Three counter-examples isolating each hypothesis of Rolle.

Continuity at the endpoints fails: $f : [0, 1] \to \mathbb{R}$, $f(x) = x$ for $x \in [0, 1[$ and $f(1) = 0$. Then $f(0) = f(1) = 0$, $f$ is differentiable on $]0, 1[$, but $f$ is not continuous at $1$; and $f'(x) = 1$ never vanishes.
Interior differentiability fails: $f(x) = |x - 1/2|$ on $[0, 1]$ --- continuous, $f(0) = f(1) = 1/2$, but not differentiable at $1/2$; and $f'$ vanishes nowhere on $]0, 1[ \setminus \{1/2\}$.
$f(a) \ne f(b)$: $f(x) = x$ on $[0, 1]$ --- $f$ is continuous and differentiable, but $f(0) = 0 \ne 1 = f(1)$, and $f'(x) = 1$ never vanishes.

Each hypothesis is necessary.

Remark --- Rolle is false over $\mathbb{C}$

Rolle does NOT extend to complex-valued functions. This counterexample will be revisited in the complex-valued extension section to explain why the equality form of the mean value theorem has no complex-valued analogue.

Theorem — Mean value theorem (equality form)

Let $a < b$ and $f : [a, b] \to \mathbb{R}$ continuous on $[a, b]$ and differentiable on $]a, b[$. Then there exists $c \in {]}a, b{[}$ such that $$ \textcolor{colorprop}{f(b) - f(a) = f'(c) (b - a)}. $$

Proof

Define the affine function $d : [a, b] \to \mathbb{R}$ corresponding to the chord: $$ d(x) = f(a) + \frac{f(b) - f(a)}{b - a} (x - a), \qquad d(a) = f(a), \quad d(b) = f(b). $$ Set $\varphi(x) = f(x) - d(x)$. Then $\varphi$ is continuous on $[a, b]$ (sum of two continuous), differentiable on $]a, b[$ with $\varphi'(x) = f'(x) - (f(b) - f(a))/(b - a)$, and $\varphi(a) = \varphi(b) = 0$. By Rolle T4.1 applied to $\varphi$, there exists $c \in {]}a, b{[}$ with $\varphi'(c) = 0$, i.e. $f'(c) = (f(b) - f(a))/(b - a)$, which rearranges to the conclusion.

Figure --- Mean value theorem

Mean value theorem: there exists an interior point $c$ where the tangent to the graph of $f$ is parallel to the chord joining $(a, f(a))$ and $(b, f(b))$.

Example

Use the mean value theorem to show that $\forall x > 0, \sqrt{x + 1} - \sqrt{x} \le \frac{1}{2 \sqrt{x}}$.

Answer

Apply the mean value theorem to $g(t) = \sqrt{t}$ on $[x, x + 1]$ (with $x > 0$): $g$ is continuous on $[x, x + 1]$, differentiable on $]x, x + 1[$ with $g'(t) = 1/(2 \sqrt{t})$. By T4.2, $\exists c \in {]}x, x + 1{[}$ with $g(x+1) - g(x) = g'(c) \cdot 1 = 1/(2 \sqrt{c})$. Since $c > x > 0$, $\sqrt{c} > \sqrt{x}$ hence $1/(2 \sqrt{c}) < 1/(2 \sqrt{x})$, so $\sqrt{x+1} - \sqrt{x} \le 1/(2 \sqrt{x})$.

Skills to practice

Applying Rolle's theorem to find a zero of $f'$
Applying the mean value theorem (equality form)

V Mean value inequality$\virgule$ monotonicity$\virgule$ and the limit of the derivative

Three consequences of the mean value theorem: the mean value inequality --- Lipschitz bound from a bound on $|f'|$; the link sign-of-$f'$ $\leftrightarrow$ monotonicity; the theorem of the limit of the derivative --- a rigorous $C^1$ extension at a tricky point. The mean value inequality + contraction $k < 1$ also yields the convergence rate of the recurrent sequences of Real sequences.

Theorem — Mean value inequality

Let $I$ be an interval and $f : I \to \mathbb{R}$ continuous on $I$, differentiable on $\mathring{I}$ (the interior of $I$). If $|f'(x)| \le K$ for all $x \in \mathring{I}$, then $f$ is $K$-Lipschitz on $I$: $$ \textcolor{colorprop}{\forall (x, y) \in I^2, \quad |f(y) - f(x)| \le K |y - x|}. $$

Proof

For $x, y \in I$, treat $x \ne y$ (if $x = y$ the inequality is $0 \le 0$). WLOG $x < y$. Then $[x, y] \subset I$, $f$ is continuous on $[x, y]$ and differentiable on $]x, y[ \subset \mathring{I}$. By the mean value theorem T4.2 applied on $[x, y]$, there exists $c \in {]}x, y{[}$ with $f(y) - f(x) = f'(c) (y - x)$. Hence $$ |f(y) - f(x)| = |f'(c)| \cdot |y - x| \le K |y - x|. $$

Method — Lipschitzianity from a bound on $f'$

To show $f$ is $K$-Lipschitz on an interval $I$:

check $f$ is continuous on $I$ and differentiable on $\mathring{I}$;
bound $|f'(x)| \le K$ for $x \in \mathring{I}$;
conclude by T5.1.

Example

Show that $f : [0, +\infty[ \to \mathbb{R}$, $f(x) = \sqrt{x + 1}$, is $(1/2)$-Lipschitz.

Answer

$f$ is continuous on $[0, +\infty[$ and differentiable on $]0, +\infty[$ with $f'(x) = 1/(2 \sqrt{x + 1})$. For $x \in ]0, +\infty[$, $x + 1 > 1$ so $\sqrt{x + 1} > 1$ and $|f'(x)| = 1/(2 \sqrt{x+1}) \le 1/2$. By T5.1, $f$ is $(1/2)$-Lipschitz.

Method — Contraction --- bridge to the recurrent sequences of Real sequences

Setup. $f : [a, b] \to [a, b]$ continuous (so $[a, b]$ is stable by $f$), differentiable on $]a, b[$, with $|f'(x)| \le k$ on $]a, b[$ for some $k \in [0, 1[$. Then $f$ has a unique fixed point $\ell \in [a, b]$, and for every $u_0 \in [a, b]$ the recurrent sequence $u_{n+1} = f(u_n)$ stays in $[a, b]$ and converges geometrically to $\ell$ with $|u_n - \ell| \le k^n |u_0 - \ell|$. Standard recipe:

Existence. Set $g(x) = f(x) - x$ on $[a, b]$. Stability gives $g(a) \ge 0$ and $g(b) \le 0$; the intermediate value theorem (Limits and continuity T6.1) yields $\ell \in [a, b]$ with $f(\ell) = \ell$.
Stability of $(u_n)$. Induction: $u_0 \in [a, b]$; if $u_n \in [a, b]$, then $u_{n+1} = f(u_n) \in f([a, b]) \subset [a, b]$.
Geometric rate. Apply T5.1 with $K = k$ on $[\min(u_n, \ell), \max(u_n, \ell)] \subset [a, b]$: $|u_{n+1} - \ell| \le k |u_n - \ell|$, hence $|u_n - \ell| \le k^n |u_0 - \ell| \to 0$.
Uniqueness. If $\ell'$ is another fixed point, taking $u_0 = \ell'$ makes the sequence constant equal to $\ell'$ while it converges to $\ell$, so $\ell' = \ell$.

Proposition — Sign of $f'$ and monotonicity

Let $I$ be an interval, $f : I \to \mathbb{R}$ continuous on $I$ and differentiable on $\mathring{I}$. Then:

(a) $f$ is constant on $I$ $\iff$ $\textcolor{colorprop}{f' \equiv 0}$ on $\mathring{I}$ ;
(b) $f$ is increasing on $I$ $\iff$ $\textcolor{colorprop}{f' \ge 0}$ on $\mathring{I}$ (symmetric for decreasing).

Proof

Prove (b). $(\Rightarrow)$ If $f$ is increasing on $I$, then for any $a \in \mathring{I}$ and any sufficiently small $h > 0$ with $a + h \in I$, $f(a + h) \ge f(a)$, hence $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h} \ge 0. $$ Passage to the limit as $h \to 0^+$ for function limits (Limits and continuity P4.1) gives $f'(a) \ge 0$.
$(\Leftarrow)$ Suppose $f' \ge 0$ on $\mathring{I}$. For $x, y \in I$ with $x < y$, $[x, y] \subset I$, $f$ is continuous on $[x, y]$ and differentiable on $]x, y[ \subset \mathring{I}$. By T4.2 (mean value theorem), there exists $c \in {]}x, y{[}$ with $f(y) - f(x) = f'(c) (y - x) \ge 0$ (since $f'(c) \ge 0$ and $y - x > 0$). Hence $f(y) \ge f(x)$, i.e. $f$ is increasing.
Deduce (a). If $f$ is constant, then $f' \equiv 0$ on $\mathring{I}$ directly. Conversely, if $f' \equiv 0$, applying (b) to $f$ gives $f$ increasing, and applying (b) to $-f$ (whose derivative is also $\equiv 0$) gives $-f$ increasing, i.e. $f$ decreasing. So $f$ is both increasing and decreasing on $I$, hence constant.

Proposition — Strict monotonicity --- useful complement

Under the same hypotheses as P5.1, $f$ is strictly increasing on $I$ $\iff$ $\textcolor{colorprop}{f' \ge 0}$ on $\mathring{I}$ and $f'$ is not identically zero on any non-trivial subinterval of $\mathring{I}$.

Proof

$(\Rightarrow)$ If $f$ is strictly increasing, then $f' \ge 0$ on $\mathring{I}$ by P5.1(b). Moreover, if $f' \equiv 0$ on some non-trivial subinterval $J \subset \mathring{I}$, then $f$ would be constant on $J$ by P5.1(a), contradicting strict monotonicity.
$(\Leftarrow)$ From $f' \ge 0$ on $\mathring{I}$ we get $f$ increasing on $I$ (P5.1(b)). Suppose for contradiction that $f(x) = f(y)$ for some $x < y$ in $I$. Since $f$ is increasing, this forces $f \equiv f(x)$ on $[x, y]$; then $f' \equiv 0$ on ${]}x, y{[}$ by P5.1(a), contradicting the hypothesis.

Remark --- $I$ must be an interval

The hypothesis « $I$ is an interval » is essential in P5.1. Counterexample: $D = \,]-\infty, 0[ \,\cup\, ]0, +\infty[ = \mathbb{R}^*$ (which is not an interval --- the same punctured domain that appears naturally when studying the Heaviside example below). Define $f : D \to \mathbb{R}$ by $f(x) = 0$ for $x < 0$ and $f(x) = 1$ for $x > 0$. Then $f' \equiv 0$ on $D$ (constant on each connected component) but $f$ is not constant on $D$.

Example

Study the variations of $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x/(1 + x^2)$.

Answer

$f$ is differentiable on $\mathbb{R}$ (rational function, denominator never zero) with $$ f'(x) = \frac{(1 + x^2) - x \cdot 2 x}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2}. $$ Sign of $f'(x)$: same sign as $1 - x^2 = (1 - x)(1 + x)$. Positive for $x \in {]}-1, 1{[}$, negative outside. By P5.1, $f$ is increasing on $[-1, 1]$ and decreasing on $]-\infty, -1]$ and $[1, +\infty[$. Maximum global at $x = 1$: $f(1) = 1/2$. Minimum global at $x = -1$: $f(-1) = -1/2$.

Theorem — Theorem of the limit of the derivative

Let $I$ be an interval, $a \in I$, $f : I \to \mathbb{R}$ continuous on $I$ and differentiable on $I \setminus \{a\}$. Suppose $f'(x) \to \ell \in \overline{\mathbb{R}}$ as $x \to a$ (with $x \in I \setminus \{a\}$). Then:

If $\ell \in \mathbb{R}$: $f$ is differentiable at $a$ and $\textcolor{colorprop}{f'(a) = \ell}$. If moreover $f'$ is continuous on $I \setminus \{a\}$, then $f' : I \to \mathbb{R}$ (extended by $f'(a) := \ell$) is continuous at $a$, hence $f$ is $C^1$ on a neighborhood of $a$ in $I$.
If $\ell = \pm \infty$: the difference quotient satisfies $\tau_a(h) \to \ell$ as $h \to 0$, and the graph of $f$ admits a vertical (half-)tangent at $a$.

Proof

Finite case $\ell \in \mathbb{R}$. Let $h \ne 0$ be such that $a + h \in I$. By the mean value theorem T4.2 applied to $f$ on the segment with endpoints $a$ and $a + h$ (continuous on the closed segment, differentiable on the open segment $\subset I \setminus \{a\}$), there exists $c_h$ strictly between $a$ and $a + h$ such that $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h} = f'(c_h). $$ As $h \to 0$, $c_h$ is squeezed between $a$ and $a + h$, hence $c_h \to a$. By hypothesis $f'(c_h) \to \ell$ (composition of limits), so $\tau_a(h) \to \ell$. Thus $f$ is differentiable at $a$ with $f'(a) = \ell$. If $a$ is an endpoint of $I$, the same argument applies with $h$ on the single side that fits inside $I$, and the conclusion holds as a one-sided derivative.
Case $\ell = \pm \infty$. The same argument gives $\tau_a(h) = f'(c_h)$ with $c_h \to a$ as $h \to 0$. Hence $\tau_a(h) \to \pm \infty$, which means the graph admits a vertical tangent at $a$ (with the convention $f'(a) = \pm \infty$, not a real derivative).

Remark --- continuity of $f$ is essential

Distinct from « extending $f'$ continuously ». The hypothesis « $f$ continuous at $a$ » in T5.2 is essential. Counterexample: the Heaviside function $H(x) = 0$ for $x < 0$ and $H(x) = 1$ for $x \ge 0$. Then $\lim_{x \to 0^-} H(x) = 0 \ne 1 = H(0)$, so $H$ is discontinuous at $0$ and the hypothesis of T5.2 fails. The naive « plug in $\lim H' = 0$ » conclusion would be wrong: although $H' \equiv 0$ on $\mathbb{R}^* = \mathbb{R} \setminus \{0\}$, hence $H'(x) \to 0$ as $x \to 0$, the difference quotient at $0$ reflects the discontinuity: $$ \tau_0(h) = \frac{H(0 + h) - H(0)}{h} = \frac{H(h) - 1}{h} \to \begin{cases} +\infty & (h \to 0^-) \\ 0 & (h \to 0^+) \end{cases} $$ so $H$ is not differentiable at $0$. Always verify continuity of $f$ at $a$ before invoking T5.2.

Method — Establishing $C^1$ at a point via the limit of the derivative

When the difference quotient is difficult to compute directly, an alternative recipe is:

check $f$ is continuous at $a$ and differentiable on $I \setminus \{a\}$;
compute $\lim_a f'$;
if the limit is finite, T5.2 gives differentiability at $a$ with $f'(a) = \lim_a f'$, and if $f'$ is also continuous on $I \setminus \{a\}$, $f$ is $C^1$ in a neighborhood of $a$.

Example

Show that $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$ for $x \le 0$ and $f(x) = x^3$ for $x > 0$ is differentiable at $0$, via M5.2.

Answer

Continuity at $0$: $\lim_{x \to 0^-} x^2 = 0 = f(0)$ and $\lim_{x \to 0^+} x^3 = 0 = f(0)$, so $f$ is continuous at $0$.
Differentiability on $\mathbb{R}^*$: $f$ is polynomial on each open half-line, hence differentiable with $f'(x) = 2x$ for $x < 0$ and $f'(x) = 3x^2$ for $x > 0$.
Limit of $f'$: $\lim_{x \to 0^-} 2x = 0$ and $\lim_{x \to 0^+} 3x^2 = 0$, so $\lim_{x \to 0} f'(x) = 0$ (finite).
Apply T5.2: the hypotheses of T5.2 hold ($f$ continuous on $\mathbb{R}$, differentiable on $\mathbb{R}^*$, $f'$ has a finite limit at $0$), so $f$ is differentiable at $0$ with $f'(0) = 0$.

Skills to practice

Showing that a function is Lipschitz from a bound on $f'$
Studying monotonicity from the sign of $f'$
Applying the contraction principle to recurrent sequences
Establishing $C^1$ at a point via the limit-of-derivative theorem

VI Higher-order derivatives and classes $C^k$

Iterating differentiation. The class $C^k$ formalises smoothness. We state and prove the Leibniz formula first (a direct calculation), then deduce the stability of $C^k$ under product as a corollary. Composition and inverse stability are admitted at this level.

Definition — $n$-th derivative

Recursive definition: $f^{(0)} = f$. If $f^{(n)}$ is defined on $I$ and differentiable on $I$, then $f^{(n+1)} = (f^{(n)})'$. The function $f$ is said to be $n$-times differentiable on $I$ if $f^{(n)}$ exists on $I$.

Definition — Classes $C^k$ and $C^\infty$

For $k \in \mathbb{N}$, $f$ is of class $C^k$ on $I$ if $f^{(k)}$ exists on $I$ and is continuous on $I$. $f$ is of class $C^\infty$ on $I$ if $f$ is $C^k$ for every $k \in \mathbb{N}$. Notation: $\textcolor{colordef}{C^k(I, \mathbb{R})}$, $\textcolor{colordef}{C^\infty(I, \mathbb{R})}$.

Example

Polynomials are $C^\infty$ on $\mathbb{R}$.

Answer

For $P(x) = x^n$, by induction $P^{(k)}(x) = n (n-1) \cdots (n - k + 1) x^{n-k}$ for $k \le n$, and $P^{(k)} \equiv 0$ for $k > n$. Each $P^{(k)}$ is a polynomial, hence continuous on $\mathbb{R}$. Same for any polynomial $\sum c_k x^k$ by linearity. Forward reference: $\exp$, $\sin$, $\cos$ are also $C^\infty$ on $\mathbb{R}$; their derivatives are rigorously justified in Standard functions.

Example

Show that $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x |x|$, is $C^1$ on $\mathbb{R}$ but not $C^2$.

Answer

$f(x) = x^2$ for $x \ge 0$, $f(x) = -x^2$ for $x < 0$. Both pieces are $C^\infty$ on their respective open half-line. At $0$: $f$ is continuous (both branches give $0$); $f'(x) = 2 x$ for $x > 0$, $f'(x) = -2 x$ for $x < 0$, both tend to $0$ as $x \to 0$, and $f$ is continuous at $0$, so by T5.2, $f$ is differentiable at $0$ with $f'(0) = 0$. On $\mathbb{R}$, $f'(x) = 2|x|$ --- continuous on $\mathbb{R}$, so $f \in C^1(\mathbb{R})$.
But $f'(x) = 2 |x|$ is not differentiable at $0$ (Ex1.3), so $f''(0)$ does not exist, and $f \notin C^2(\mathbb{R})$. (On $\mathbb{R}^*$, $f \in C^\infty$.)

Theorem — Leibniz formula

Let $f, g$ be $n$ times differentiable on $I$. Then $f g$ is $n$ times differentiable on $I$ and $$ \textcolor{colorprop}{(f g)^{(n)} = \sum_{p = 0}^{n} \binom{n}{p} f^{(p)} g^{(n - p)}}. $$

Proof

Induction on $n$.

Base $n = 0$. $(f g)^{(0)} = f g = \binom{0}{0} f^{(0)} g^{(0)}$, formula holds.
Inductive step. Assume the formula holds at rank $n$ for all $n$-times differentiable functions, and let $f, g$ be $(n+1)$-times differentiable. In particular $f, g$ are $n$-times differentiable, so $(f g)^{(n)} = \sum_{p = 0}^{n} \binom{n}{p} f^{(p)} g^{(n - p)}$. Differentiate once more, using linearity (P2.1) and the product rule (P2.2): $$ \begin{aligned} (f g)^{(n+1)} &= \sum_{p = 0}^{n} \binom{n}{p} \big( f^{(p+1)} g^{(n - p)} + f^{(p)} g^{(n - p + 1)} \big) \\ &= \underbrace{\sum_{p = 0}^{n} \binom{n}{p} f^{(p+1)} g^{(n - p)}}_{S_1} + \underbrace{\sum_{p = 0}^{n} \binom{n}{p} f^{(p)} g^{(n - p + 1)}}_{S_2}. \end{aligned} $$ Re-index $S_1$ with $q = p + 1$ (so $q$ runs from $1$ to $n + 1$); rename $q$ as $p$: $$ S_1 = \sum_{p = 1}^{n+1} \binom{n}{p - 1} f^{(p)} g^{(n + 1 - p)}. $$ Keep $S_2$ as it is. Adding $S_1 + S_2$, the $p = 0$ term comes from $S_2$ alone (coefficient $\binom{n}{0} = 1 = \binom{n+1}{0}$), the $p = n + 1$ term from $S_1$ alone (coefficient $\binom{n}{n} = 1 = \binom{n+1}{n+1}$), and for $1 \le p \le n$ Pascal's relation $\binom{n}{p - 1} + \binom{n}{p} = \binom{n + 1}{p}$ (chapter Sums, products and binomial coefficients) groups the two contributions. Hence $$ (f g)^{(n+1)} = \sum_{p = 0}^{n+1} \binom{n+1}{p} f^{(p)} g^{(n + 1 - p)}, $$ which is the formula at rank $n + 1$.

Proposition — Stability of $C^k$

Let $f, g \in C^k(I, \mathbb{R})$, $\lambda, \mu \in \mathbb{R}$. Then:

$\textcolor{colorprop}{\lambda f + \mu g \in C^k(I)}$ ;
$\textcolor{colorprop}{f g \in C^k(I)}$ ;
if $g$ does not vanish on $I$, $\textcolor{colorprop}{f / g \in C^k(I)}$ ;
if $\varphi : J \to I$ is $C^k$, then $\textcolor{colorprop}{f \circ \varphi \in C^k(J)}$ ;
if $k \ge 1$, $f \in C^k(I, \mathbb{R})$ with $f : I \to f(I)$ bijective and $f'$ not vanishing on $I$, then $\textcolor{colorprop}{f^{-1} \in C^k(f(I), \mathbb{R})}$.

Proof

Linear combination. Induction on $k$. Base $k = 0$: $\lambda f + \mu g$ is continuous (sum of continuous functions). Inductive step $k \to k + 1$: if $f, g$ are $C^{k+1}$, they are differentiable and $f', g'$ are $C^k$; by P2.1 $(\lambda f + \mu g)' = \lambda f' + \mu g'$ is $C^k$ by induction, hence $\lambda f + \mu g$ is $C^{k+1}$.
Product. Same induction. Base $k = 0$: $f g$ is continuous. Inductive step: $f, g$ are $C^{k+1}$, so $(f g)' = f' g + f g'$ (P2.2). Each summand is a product of $C^k$ functions, hence $C^k$ by induction; their sum is $C^k$ by linearity; thus $f g$ is $C^{k+1}$.
Quotient. First show by induction on $k$ that if $g$ is $C^k$ and does not vanish on $I$, then $1/g$ is $C^k$ on $I$. Base $k = 0$: $1/g$ is continuous (continuity of $g$ and non-vanishing). Inductive step: if $g$ is $C^{k+1}$, then $(1/g)' = -g'/g^2$ (P2.3); by the product case $g^2$ is $C^k$, $g^2$ does not vanish, so by the induction hypothesis $1/g^2$ is $C^k$, and $-g'/g^2 = -g' \cdot (1/g^2)$ is $C^k$ as a product; hence $1/g$ is $C^{k+1}$. Then $f/g = f \cdot (1/g)$ is $C^k$ as a product.
Composition and inverse: proofs admitted. The proofs concerning composition and inverse are not required at this level. The results themselves remain in scope and may be used freely.

Method — Compute a higher-order derivative

Three classical patterns:

Pattern A --- $P(x) \cdot \exp(a x)$. Apply Leibniz; only the first $\deg P + 1$ terms are non-zero since $P^{(k)} = 0$ for $k > \deg P$.
Pattern B --- rational fraction. Decompose into partial fractions (chapter Rational fractions), then use $(1/(x - a))^{(n)} = (-1)^n n! / (x - a)^{n + 1}$ --- proved by direct induction.
Pattern C --- trig power. Linearise first (chapter Trigonometry), then differentiate the linearised expression term by term.

Admitted iterated derivatives (forward references). $(\exp(a x))^{(k)} = a^k \exp(a x)$ on $\mathbb{R}$ for all $k \in \mathbb{N}$, admitted from Real functions: lycée recap. Iterated derivatives of $\sin$, $\cos$ are admitted from Standard functions.

Example

Compute $\big(1/(x^2 - 1)\big)^{(n)}$ on $\mathbb{R} \setminus \{-1, 1\}$ via Pattern B.

Answer

Partial fractions: $1/(x^2 - 1) = 1/((x - 1)(x + 1)) = (1/2) \big(1/(x - 1) - 1/(x + 1)\big)$. By the identity $(1/(x - a))^{(n)} = (-1)^n n! / (x - a)^{n + 1}$ (induction), with linearity (P2.1 extended to higher orders): for every $x \in \mathbb{R} \setminus \{-1, 1\}$ (equivalently, on each interval of the domain), $$ \left(\frac{1}{x^2 - 1}\right)^{(n)} = \frac{(-1)^n n!}{2} \left( \frac{1}{(x - 1)^{n + 1}} - \frac{1}{(x + 1)^{n + 1}} \right). $$

Example

Compute $\big( (x^2 - 1) \exp(x) \big)^{(5)}$ via Pattern A.

Answer

Set $f(x) = x^2 - 1$, $g(x) = \exp(x)$. Derivatives: $f' = 2 x$, $f'' = 2$, $f^{(k)} = 0$ for $k \ge 3$. $g^{(k)} = \exp(x)$ for all $k$. By Leibniz with $n = 5$: $$ \begin{aligned} (f g)^{(5)} &= \binom{5}{0} f^{(0)} g^{(5)} + \binom{5}{1} f^{(1)} g^{(4)} + \binom{5}{2} f^{(2)} g^{(3)} \\ &\quad + \binom{5}{3} f^{(3)} g^{(2)} + \binom{5}{4} f^{(4)} g^{(1)} + \binom{5}{5} f^{(5)} g^{(0)} \\ &= (x^2 - 1) \exp(x) + 5 \cdot 2 x \cdot \exp(x) + 10 \cdot 2 \cdot \exp(x) \\ &\quad + 0 + 0 + 0 \\ &= (x^2 + 10 x + 19) \exp(x). \end{aligned} $$

Skills to practice

Computing higher-order derivatives via Leibniz
Determining the class $C^k$ of a piecewise function

VII Brief extension to complex-valued functions

Differentiability extends to functions $f : I \to \mathbb{C}$ via componentwise differentiability. The algebraic toolbox carries over almost verbatim. The crucial obstruction: Rolle's theorem and the equality form of the mean value theorem are false over $\mathbb{C}$. However the mean value inequality for $C^1$ functions still holds --- admitted here, the proof being deferred to Integration on a segment.

Definition — Complex-valued differentiability

$f : I \to \mathbb{C}$, $a \in I$. We say $f$ is differentiable at $a$ if the difference quotient $$ \tau_a(h) = \frac{f(a + h) - f(a)}{h} $$ admits a finite limit in $\mathbb{C}$ as $h \to 0$ with $a + h \in I$. The limit is denoted $\textcolor{colordef}{f'(a)}$.

Proposition — Componentwise characterization

$f : I \to \mathbb{C}$ is differentiable at $a \in I$ if and only if $\mathrm{Re}\,f$ and $\mathrm{Im}\,f$ are both differentiable at $a$. In this case, $$ \textcolor{colorprop}{f'(a) = (\mathrm{Re}\,f)'(a) + i \, (\mathrm{Im}\,f)'(a)}. $$

Proof

For $h \ne 0$ with $a + h \in I$, $$ \tau_a^f(h) = \tau_a^{\mathrm{Re}\,f}(h) + i \, \tau_a^{\mathrm{Im}\,f}(h) $$ by $\mathbb{R}$-linearity of $\mathrm{Re}$ and $\mathrm{Im}$. By the componentwise characterization of complex-valued function limits (Limits and continuity, complex-valued functions section), $\tau_a^f(h)$ has a finite limit in $\mathbb{C}$ as $h \to 0$ iff $\tau_a^{\mathrm{Re}\,f}$ and $\tau_a^{\mathrm{Im}\,f}$ both have finite limits in $\mathbb{R}$, and the limit decomposes as $(\mathrm{Re}\,f)'(a) + i (\mathrm{Im}\,f)'(a)$.

Example

Compute the derivative of $f : \mathbb{R} \to \mathbb{C}$, $f(t) = \exp(i t)$.

Answer

$\mathrm{Re}\,f(t) = \cos t$, $\mathrm{Im}\,f(t) = \sin t$. Both are differentiable on $\mathbb{R}$ with $(\cos)'(t) = -\sin t$ and $(\sin)'(t) = \cos t$ (admitted from Real functions: lycée recap). By P7.1, $f$ is differentiable on $\mathbb{R}$ with $$ f'(t) = -\sin t + i \cos t = i (\cos t + i \sin t) = i \exp(i t). $$

Proposition — What does NOT change

Let $f, g : I \to \mathbb{C}$ differentiable on $I$, $\lambda, \mu \in \mathbb{C}$. Then:

$\lambda f + \mu g$ is differentiable on $I$ with $\textcolor{colorprop}{(\lambda f + \mu g)' = \lambda f' + \mu g'}$ ;
$f g$ is differentiable on $I$ with $\textcolor{colorprop}{(f g)' = f' g + f g'}$ ;
if $g$ does not vanish on $I$, $f/g$ is differentiable with $\textcolor{colorprop}{(f/g)' = (f' g - f g')/g^2}$ ;
if $\varphi : J \to I$ is a real-valued differentiable function, $f \circ \varphi$ is differentiable with $\textcolor{colorprop}{(f \circ \varphi)' = (f' \circ \varphi) \cdot \varphi'}$.

Each is deduced from the real case by componentwise decomposition via P7.1.

Proposition — What DOES change

Rolle's theorem and the equality form of the mean value theorem are false for complex-valued functions.

Counter-example

The function $f : [0, 2 \pi] \to \mathbb{C}$, $f(t) = \exp(i t)$, satisfies $f(0) = f(2 \pi) = 1$ and is $C^\infty$ on $[0, 2 \pi]$ with $f'(t) = i \exp(i t)$ (Ex7.1). But $|f'(t)| = 1$ everywhere, so $f'$ never vanishes --- Rolle fails. As a consequence, no value of $c \in {]}0, 2 \pi[$ satisfies $f(2\pi) - f(0) = f'(c) \cdot 2 \pi$ (the left side is $0$, the right side has modulus $2 \pi$), so the equality form of the mean value theorem fails too.

Theorem — Mean value inequality for complex-valued $C^1$ functions

Let $f : I \to \mathbb{C}$ of class $C^1$ on $I$ with $|f'(x)| \le K$ for all $x \in I$. Then for all $(x, y) \in I^2$, $$ \textcolor{colorprop}{|f(y) - f(x)| \le K |y - x|}. $$

Remark --- admitted here

T7.1 is admitted in this chapter. The inequality results from a simple integral upper bound, justified later in the Integration section. The full proof will be given in Integration on a segment (via $f(y) - f(x) = \int_x^y f'(t) dt$ and the triangle inequality for integrals). Until then, T7.1 is used as a black box.

Example

Use T7.1 to show $|\exp(i t) - 1| \le |t|$ for all $t \in \mathbb{R}$.

Answer

Set $f(t) = \exp(i t)$ on $\mathbb{R}$. Then $f$ is $C^1$ on $\mathbb{R}$ with $f'(t) = i \exp(i t)$ (Ex7.1), so $|f'(t)| = |i| \cdot |\exp(i t)| = 1$ for all $t$. By T7.1 with $K = 1$, $|f(t) - f(0)| \le 1 \cdot |t - 0|$, i.e. $|\exp(i t) - 1| \le |t|$.

Skills to practice

Differentiating a complex-valued function componentwise
Applying the complex mean value inequality