Differentiability

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

:::caution Common Pitfall L’Hôpital’s rule only applies to indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$ . 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. Also, L’Hôpital’s rule requires that the Limit of the quotient of derivatives exists; if it does not exist (oscillates), the original limit May still exist.

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