Differentiability | Mathematics

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 $0$.

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

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

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.

Corollary 5.6. If $f$ is differentiable on $(a,b)$ and $|f'(x)| \leq M$ for all $x \in (a,b)$ Then $f$ is Lipschitz continuous with constant $M$: $|f(x) - f(y)| \leq M|x - y|$ for all $x, y \in (a,b)$.

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

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 $[a, x]$. $\blacksquare$

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 $f(x)/g(x) \to L$. $\blacksquare$

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. 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 $x$.

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.

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.

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

• L’Hopital’s rule only applies to indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Applying it to forms like $\frac{1}{0}$ or $\frac{\infty}{1}$ will give incorrect results. Always verify the indeterminate form before applying the rule.
• L’Hopital’s rule requires that the limit of $f'/g'$ exists. If $f'/g'$ oscillates (e.g., $f(x) = x + \sin x$, $g(x) = x$, then $f'/g' = 1 + \cos(x)/1$ which oscillates), the original limit may still exist. In such cases, use algebraic manipulation instead.
• The MVT requires continuity on $[a, b]$ and differentiability on $(a, b)$. If $f$ is not differentiable at some points, the MVT conclusion may fail.
• Taylor’s theorem gives the remainder for a specific $\xi$ between $a$ and $x$. To bound the error, use $|R_n(x)| \leq \frac{M}{(n+1)!}|x - a|^{n+1}$ where $M = \sup_{\xi}|f^{(n+1)}(\xi)|$.

flowchart TD
    A[5_Differentiability] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

• Continuity: Differentiability implies continuity, and the mean value theorem connects local derivative behaviour to global function properties.
• Riemann Integration: The fundamental theorem of calculus links differentiation and integration, with the Riemann integral defined via limits of Riemann sums.
• Lebesgue Integration: Lebesgue integration generalises the Riemann integral and handles a broader class of functions with discontinuities.
• Classical Mechanics
• Electromagnetism