Differentiability

Recall prompts

Attempt these from memory before reading.

Define the derivative as a limit. What kind of limit is it?
State the mean value theorem. What are its hypotheses, and which one is load-bearing?
Is every continuous function differentiable? Is every differentiable function continuous?
Can a derivative be discontinuous?
State Taylor’s theorem with the Lagrange form of the remainder.

Check your answers against the text. Prompt 4 is the one most people answer wrongly, and Darboux’s theorem is the reason the right answer is surprising.

Motivation: the derivative is a linear map, not a slope

“Slope of the tangent line” is how differentiation is introduced, and it is a useful lie. The honest definition is:

$f$ is differentiable at $a$ if there is a number $f'(a)$ with $f(a + h) = f(a) + f'(a) h + o(h)$ .

That is: $f$ is differentiable at $a$ exactly when it can be approximated near $a$ by an affine function, with error smaller than linear. The number $f'(a)$ is the slope of that affine approximation.

Reading it this way does three things the slope picture cannot:

It generalises to $\mathbb{R}^n$ , where the derivative is a linear map (the Jacobian) and there is no single slope.
It makes the mean value theorem obvious: if the function is affine on average, some point must have the affine slope.
It explains why differentiability is stronger than continuity. Continuity says $f(a+h) - f(a) \to 0$ ; differentiability says the difference is linear in $h$ . Linear is much stronger.

The counterexamples on this page are the reason the distinction has to be made carefully rather than assumed.

5.1 The Derivative

Definition. $f : (a,b) \to \mathbb{R}$ is differentiable at $c \in (a,b)$ if the limit

f"(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}

Exists (as a finite real number).

Proposition 5.1. If $f$ is differentiable at $c$ Then $f$ is continuous at $c$ .

Proof. $\lim_{x \to c} (f(x) - f(c)) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \cdot (x - c) = f'(c) \cdot 0 = 0$ . $\blacksquare$

The converse is false: $f(x) = |x|$ is continuous at $0$ but not differentiable at $0$ .

5.2 Differentiation Rules

Theorem 5.1. If $f$ and $g$ are differentiable at $c$ Then:

$(f + g)'(c) = f'(c) + g'(c)$
$(fg)'(c) = f'(c)g(c) + f(c)g'(c)$
$(f/g)'(c) = \frac{f'(c)g(c) - f(c)g'(c)}{g(c)^2}$ (if $g(c) \neq 0$ )
$(f \circ g)'(c) = f'(g(c)) \cdot g'(c)$ (Chain Rule)

5.3 Mean Value Theorem

Theorem 5.2 (Rolle’s Theorem). If $f : [a,b] \to \mathbb{R}$ is continuous on $[a,b]$ Differentiable On $(a,b)$ And $f(a) = f(b)$ Then there exists $c \in (a,b)$ such that $f'(c) = 0$ .

Proof. By the Extreme Value Theorem, $f$ attains its maximum $M$ and minimum $m$ on $[a,b]$ . If $M = m$ Then $f$ is constant and $f'(c) = 0$ for all $c \in (a,b)$ . Otherwise, at least one Of $M$ or $m$ is attained at some $c \in (a,b)$ (since $f(a) = f(b)$ ). By Fermat’s theorem, $f'(c) = 0$ . $\blacksquare$

Theorem 5.3 (Mean Value Theorem). If $f : [a,b] \to \mathbb{R}$ is continuous on $[a,b]$ and Differentiable on $(a,b)$ Then there exists $c \in (a,b)$ such that

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

Proof. Define $g(x) = f(x) - \frac{f(b)-f(a)}{b-a}(x - a)$ . Then $g(a) = g(b)$ and $g$ satisfies the Hypotheses of Rolle’s theorem. So $g'(c) = 0$ for some $c \in (a,b)$ Which gives the result. $\blacksquare$

Corollary 5.4. If $f'(x) = 0$ for all $x \in (a,b)$ Then $f$ is constant on $[a,b]$ .

Corollary 5.5. If $f'(x) > 0$ for all $x \in (a,b)$ Then $f$ is strictly increasing on $[a,b]$ .

Theorem 5.3a (Cauchy’s Mean Value Theorem). If $f, g : [a,b] \to \mathbb{R}$ are continuous on $[a,b]$ and differentiable on $(a,b)$ Then there exists $c \in (a,b)$ such that

(f(b) - f(a))g'(c) = (g(b) - g(a))f'(c)

Proof. Define $h(x) = (f(b) - f(a))g(x) - (g(b) - g(a))f(x)$ . Then $h(a) = h(b)$ So by Rolle’s Theorem, $h'(c) = 0$ for some $c \in (a,b)$ Which gives the result. $\blacksquare$

Remark. When $g(x) = x$ Cauchy’s MVT reduces to the standard MVT. Cauchy’s MVT is the key Ingredient in the …/1-number-and-algebra/3_proof-and-logic of L’Hôpital’s rule.

Proof. Apply the MVT to $f$ on the interval between $x$ and $y$ . $\blacksquare$

5.4 Taylor’s Theorem

Theorem 5.6 (Taylor’s Theorem with Lagrange Remainder). If $f$ is $(n+1)$ -times differentiable on An open interval containing $a$ Then for each $x$ in that interval:

f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k + R_n(x)

Where the remainder is

R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x - a)^{n+1}

For some $\xi$ between $a$ and $x$ .

Proof. Fix $x \neq a$ and define

g(t) = f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(t)}{k!}(x - t)^k

Then $g(a) = R_n(x)$ and $g(x) = 0$ . By the generalized Rolle’s theorem (or direct computation Using the Cauchy mean value theorem), there exists $\xi$ between $a$ and $x$ with $g'( \xi ) = 0$ . Computing:

g'(t) = -\frac{f^{(n+1)}(t)}{n!}(x - t)^n

Setting $g'(\xi) = 0$ yields the result after comparing $g(a) = R_n(x)$ with the integral form. A Cleaner approach uses the standard MVT applied to $g$ on $[a, x]$ . $\blacksquare$

5.5 L’Hôpital’s Rule

Theorem 5.7 (L’Hôpital’s Rule, $\frac{0}{0}$ case). Suppose $f$ and $g$ are differentiable on An open interval containing $c$ (except possibly at $c$ itself), $g'(x) \neq 0$ near $c$ And $\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0$ . If $\lim_{x \to c} f'(x)/g'(x) = L$ exists (as a finite Number or $\pm\infty$ ), then $\lim_{x \to c} f(x)/g(x) = L$ .

Proof. Extend $f$ and $g$ continuously to $c$ by setting $f(c) = g(c) = 0$ . For $x \neq c$ By Cauchy’s Mean Value Theorem, there exists $\xi$ strictly between $c$ and $x$ such that

\frac{f(x) - f(c)}{g(x) - g(c)} = \frac{f'(\xi)}{g'(\xi)}

I.e., $\frac{f(x)}{g(x)} = \frac{f'(\xi)}{g'(\xi)}$ . As $x \to c$ We have $\xi \to c$ (since $\xi$ is trapped between $c$ and $x$ ). Therefore $\lim_{x \to c} f(x)/g(x) = \lim_{\xi \to c} f'(\xi)/g'(\xi) = L$ . $\blacksquare$

Theorem 5.7b (L’Hôpital’s Rule, $\frac{\infty}{\infty}$ case). Suppose $f$ and $g$ are Differentiable on $(a, b)$ (except possibly at $c$ ), $g'(x) \neq 0$ near $c$ And $\lim_{x \to c} |f(x)| = \lim_{x \to c} |g(x)| = \infty$ . If $\lim_{x \to c} f'(x)/g'(x) = L$ exists, Then $\lim_{x \to c} f(x)/g(x) = L$ .

Proof (sketch). Fix $\varepsilon > 0$ . For $x, y$ near $c$ with $x \neq y$ By Cauchy’s MVT:

\frac{f(x) - f(y)}{g(x) - g(y)} = \frac{f'(\xi)}{g'(\xi)}

For some $\xi$ between $x$ and $y$ . Since $f'(\xi)/g'(\xi) \approx L$ for $\xi$ near $c$ We have:

\frac{f(x)}{g(x)} = \frac{f(x) - f(y)}{g(x) - g(y)} \cdot \frac{1 - f(y)/f(x)}{1 - g(y)/g(x)}

Since $f(x), g(x) \to \infty$ By fixing $y$ and letting $x \to c$ The fractions $f(y)/f(x)$ and $g(y)/g(x)$ tend to $0$ So the second factor tends to $1$ . The first factor tends to $L$ by Cauchy’s MVT. Hence $f(x)/g(x) \to L$ . $\blacksquare$

Worked Example: Compute $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

Solution. Both numerator and denominator approach $0$ as $x \to 0$ . Applying L’Hôpital’s rule:

\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x}

This is still $\frac{0}{0}$ So apply L’Hôpital again:

= \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}

$\blacksquare$

5.6 Darboux’s Theorem

Theorem 5.8 (Darboux’s Theorem). If $f$ is differentiable on $[a, b]$ Then $f'$ has the Intermediate value property: for any $y$ between $f'(a)$ and $f'(b)$ There exists $c \in (a, b)$ With $f'(c) = y$ .

Remark. This means derivatives satisfy the intermediate value property even though they need not be continuous. For example, $f(x) = x^2 \sin(1/x)$ (with $f(0) = 0$ ) is differentiable everywhere, But $f'$ is not continuous at $0$ .

Proof. Assume without loss of generality that $f'(a) \lt y \lt f'(b)$ . Define $g(x) = f(x) - yx$ . Then $g$ is differentiable on $[a, b]$ with

g'(a) = f'(a) - y \lt 0 \quad \mathrm{and} \quad g'(b) = f'(b) - y > 0

Since $g'(a) \lt 0$ There exists $x_1 > a$ with $g(x_1) \lt g(a)$ (otherwise $g(x) \geq g(a)$ For $x$ near $a$ Contradicting $g'(a) \lt 0$ ). Similarly, since $g'(b) > 0$ There exists $x_2 \lt b$ with $g(x_2) \lt g(b)$ .

Therefore $g$ attains its minimum at some $c \in (a, b)$ . By Fermat’s theorem on interior extrema, $g'(c) = 0$ So $f'(c) = y$ . $\blacksquare$

Worked Example: Apply Darboux's theorem to $f(x) = x^2 \sin(1/x)$ ($f(0) = 0$)

Solution. For $x \neq 0$ : $f'(x) = 2x \sin(1/x) - \cos(1/x)$ . At $x = 0$ : $f'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin(1/h) = 0$ .

So $f'(0) = 0$ . For any $\delta > 0$ The term $-\cos(1/x)$ oscillates between $-1$ and $1$ on $(0, \delta)$ So $f'$ takes all values in $[-1, 1]$ infinitely often on $(0, \delta)$ .

But Darboux’s theorem says $f'$ has the intermediate value property. Indeed, $f'$ is not continuous at $0$ (it oscillates wildly), yet it still satisfies the IVP. This shows that derivatives can be Highly discontinuous while retaining the intermediate value property. $\blacksquare$

5.7 Worked Examples

Worked Example. Compute the third-order Taylor polynomial of $f(x) = e^x$ about $a = 0$ .

$f(0) = 1$ , $f'(0) = 1$ , $f''(0) = 1$ , $f'''(0) = 1$ . So

T_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}

The remainder is $R_3(x) = \frac{e^\xi}{24} x^4$ for some $\xi$ between $0$ and $x$ .

Worked Example: Approximate $\sin(0.1)$ with error less than $10^{-10}$

Solution. For $f(x) = \sin x$ about $a = 0$ : $f^{(k)}(0) \in \{0, 1, -1\}$ and the Taylor Polynomial of degree $n$ has the form $T_n(x) = x - x^3/3! + x^5/5! - \cdots$ (odd terms only).

The Lagrange remainder is $|R_n(x)| = \frac{|f^{(n+1)}(\xi)|}{(n+1)!} |x|^{n+1} \leq \frac{|x|^{n+1}}{(n+1)!}$ (since $|f^{(k)}| \leq 1$ for all $k$ ).

We need $\frac{(0.1)^{n+1}}{(n+1)!} \lt 10^{-10}$ . Testing: for $n = 5$ $\frac{(0.1)^6}{6!} = \frac{10^{-6}}{720} \approx 1.39 \times 10^{-9} \gt 10^{-10}$ . For $n = 7$ : $\frac{(0.1)^8}{8!} = \frac{10^{-8}}{40320} \approx 2.48 \times 10^{-13} \lt 10^{-10}$ .

So $T_7(0.1) = 0.1 - \frac{(0.1)^3}{6} + \frac{(0.1)^5}{120} - \frac{(0.1)^7}{5040}$ $= 0.1 - 0.00016667 + 0.00000083 - 0.00000000 \approx 0.09983342$ .

The error is at most $2.48 \times 10^{-13} \lt 10^{-10}$ . $\blacksquare$

Worked Example: Find the Maclaurin series for $\ln(1 + x)$ and its radius of convergence

Solution. For $f(x) = \ln(1+x)$ : $f(0) = 0$ , $f'(x) = 1/(1+x)$ , $f''(x) = -1/(1+x)^2$ $f^{(k)}(x) = (-1)^{k-1}(k-1)!/(1+x)^k$ for $k \geq 1$ . So $f^{(k)}(0) = (-1)^{k-1}(k-1)!$ .

\ln(1+x) = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}(k-1)!}{k!} x^k = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} x^k

By the ratio test: $\lim_{k \to \infty} |a_{k+1}/a_k| = \lim_{k \to \infty} \frac{k}{k+1} |x| = |x|$ . The series converges for $|x| \lt 1$ and diverges for $|x| > 1$ . At $x = 1$ we get the alternating Harmonic series (converges to $\ln 2$ ). At $x = -1$ we get the negative harmonic series (diverges).

The radius of convergence is $R = 1$ and the interval of convergence is $(-1, 1]$ . $\blacksquare$

Worked Example: Compute the Taylor expansion of $\cos x$ about $a = \pi/3$ with remainder bound

Solution. Compute derivatives: $f(x) = \cos x$ , $f'(x) = -\sin x$ , $f''(x) = -\cos x$ , $f'''(x) = \sin x$ $f^{(4)}(x) = \cos x$ . Evaluated at $a = \pi/3$ :

$f(\pi/3) = 1/2$ , $f'(\pi/3) = -\sqrt{3}/2$ , $f''(\pi/3) = -1/2$ , $f'''(\pi/3) = \sqrt{3}/2$ .

The third-degree Taylor polynomial is:

T_3(x) = \frac{1}{2} - \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{3}\right) - \frac{1}{4}\left(x - \frac{\pi}{3}\right)^2 + \frac{\sqrt{3}}{12}\left(x - \frac{\pi}{3}\right)^3

The remainder satisfies $|R_3(x)| \leq \frac{|x - \pi/3|^4}{24}$ (since $|f^{(4)}(\xi)| = |\cos \xi| \leq 1$ ).

For example, at $x = 1$ : $|R_3(1)| \leq \frac{|1 - \pi/3|^4}{24} \approx \frac{0.0472^4}{24} \approx 2.1 \times 10^{-7}$ . $\blacksquare$

Intuition

The derivative is most deeply understood not as a “slope” but as the best linear approximation to a function at a point. Near x=a, a differentiable function f(x) is well approximated by the tangent line f(a) + f’(a)(x-a). The error in this approximation vanishes faster than |x-a| as x approaches a. This is why differentiation is local: the derivative captures the linear part of how f behaves near a, discarding higher-order curvature.

This perspective explains why the chain rule works as it does. If f is locally linear near g(x) and g is locally linear near a, then their composition is locally linear near a, and the slopes multiply. It also explains why not every continuous function is differentiable: f(x)=|x| is continuous at 0 but has no single linear approximation there — the left and right slopes disagree. The Mean Value Theorem says that the instantaneous linear approximation (the derivative) must at some point match the average rate of change over an interval. Taylor’s theorem extends this: the derivative is the first-order term in a polynomial approximation, and the remainder controls how well that approximation works.

Counterexamples worth knowing by name

Function	Continuous?	Differentiable?	What it shows
$\\|x\\|$ at $0$	yes	no	a corner: left and right derivatives differ
$x^2 \sin(1/x^2)$ , $f(0)=0$	yes	yes at 0, $f'(0)=0$	the derivative exists but is unbounded near 0
Weierstrass $W(x)$	yes everywhere	nowhere	continuity does not imply differentiability, on an uncountable set
$f(x) = x^{1/3}$ at $0$	yes	no	a vertical tangent is not a derivative
$f(x) = \begin{cases} x^2 & x \in \mathbb{Q} \\ 0 & \text{else} \end{cases}$	continuous only at 0	differentiable at 0 with $f'(0)=0$	differentiability can hold at an isolated point of continuity

Darboux’s theorem says that even though derivatives need not be continuous, they satisfy the intermediate value property: if $f'(a) < c < f'(b)$ then $f'(x) = c$ for some $x$ between $a$ and $b$ . So a derivative cannot jump. It can, however, be discontinuous — $x^2\sin(1/x^2)$ is the standard example — which is why derivatives form a strictly smaller class than continuous functions, and why “the derivative of a differentiable function is continuous” is false.

Why Weierstrass matters beyond the puzzle. It shows that continuity alone gives no control over oscillation at the infinitesimal scale. Any argument that assumes a continuous function has a derivative “somewhere” on an interval needs more hypotheses — and the mean value theorem supplies exactly that.

Interleaving problems

Mix these with material from Sequences and Limits, Continuity and Riemann Integration.

(Limits.) Show $f(x) = |x|$ is not differentiable at $0$ , by computing the left and right limits of the difference quotient.
(Continuity.) Show that if $f$ is differentiable on $[a,b]$ with $f'(a) = f'(b)$ , there is a point where $f'(c) = 0$ — this is Rolle’s theorem, and the MVT is its generalisation.
(Integration.) Use the MVT to show that if $f' \ge 0$ on $[a,b]$ then $f$ is increasing, and deduce that a function with $f' = 0$ everywhere is constant.
(Taylor.) Show that $\sin x = x - x^3/6 + o(x^4)$ and use it to compute $\lim_{x \to 0} \frac{\sin x - x}{x^3}$ .
(Darboux.) Show that $f'(x) = 2x \sin(1/x) - \cos(1/x)$ for $x e 0$ , $f'(0) = 0$ has the intermediate value property but is discontinuous at $0$ .

5.8a Geometric Meaning of the MVT

The Mean Value Theorem says: for a smooth curve connecting two points, there is a point on the curve where the tangent line is parallel to the chord (secant line) connecting the endpoints.

More precisely, if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then there exists $c \in (a, b)$ with

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

Physical interpretation: If you drive 100 miles in 2 hours, your average speed is 50 mph. The MVT says that at some instant, your speedometer must have read exactly 50 mph.

Why this fails without differentiability: For $f(x) = |x|$ on $[-1, 1]$ , $f(1) - f(-1) = 0$ , so the average rate of change is 0. But $f'(x)$ is never 0: it equals $1$ for $x > 0$ and $-1$ for $x < 0$ . The MVT fails because $f$ is not differentiable at $x = 0$ , where the chord from $(-1, 1)$ to $(1, 1)$ is horizontal, but no tangent line is horizontal.

5.8b Worked Example: MVT Inequality

Problem. Prove that $|\sin x - \sin y| \leq |x - y|$ for all $x, y \in \mathbb{R}$ .

Solution

Apply the MVT to $f(t) = \sin t$ on the interval between $x$ and $y$ . There exists $\xi$ between $x$ and $y$ such that

\sin x - \sin y = \cos(\xi) \cdot (x - y)

Taking absolute values:

|\sin x - \sin y| = |\cos(\xi)| \cdot |x - y| \leq 1 \cdot |x - y| = |x - y|

since $|\cos(\xi)| \leq 1$ for all $\xi$ .

This inequality implies that $\sin x$ is Lipschitz continuous with constant 1, hence uniformly continuous on $\mathbb{R}$ . The same argument works for $\cos x$ .