Differentiability

At a Glance

• Frequency: 3 sub-parts across 3 of 14 years (2016, 2019, 2025)
• Priority tier: T3
• Marks (count): 10 (2), 15 (1)
• Average solve time: ~9 min
• Difficulty mix: easy 3
• Section: A | Dominant type: proof

Why This Chapter Matters

Differentiability questions span two distinct archetypes — testing differentiability at a specific point using one-sided derivatives or first principles, and deriving a derivative from a functional equation. Both types appear in Section A or early Section B and carry 10–15 marks. The methods are clean and repeatable: master the first-principles difference-quotient argument and the squeeze theorem trick for oscillating functions, and you can handle every past variant. The functional-equation archetype also tests ODE fluency, giving it double value.

Minimum Theory

Differentiability. $f$ is differentiable at $x=a$ if $\displaystyle f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ exists. Equivalently, the left-hand derivative $f'_-(a)=\lim_{h\to0^-}\frac{f(a+h)-f(a)}{h}$ and right-hand derivative $f'_+(a)=\lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}$ both exist and are equal. Differentiability implies continuity; the converse is false.

Key techniques. (i) First-principles at a corner/patch point: write $g'(0)=\lim_{h\to0}g(h)/h$ and bound using $|h\sin(1/h)|\le|h|$ (squeeze theorem). (ii) Piecewise absolute-value functions: remove $|\cdot|$ separately for $x>a$ and $x<a$ (where the sign of the expression is determined), differentiate each branch, and compare the one-sided derivatives. (iii) Functional equations: write $f'(x)$ using the definition, apply the functional equation to factor out $f(x)$, and recognise the remaining limit as $f'(0)$.

Differentiable extension. To extend $f:(0,\infty)\to\mathbb R$ to a differentiable $g:\mathbb R\to\mathbb R$, set $g(x)=f(x)$ for $x>0$ and choose $g(0)$ and $g|_{x<0}$ to make both continuity and the first-principles derivative work at $0$. For $x^2\sin(1/x)$ the choice $g(0)=0$ works; the formula $g'(x)=2x\sin(1/x)-\cos(1/x)$ is valid for $x\ne 0$ but oscillates at $0$ — $g'(0)$ must be computed separately from the definition.

Question Archetypes

differentiability-test (2 question(s); 2016, 2019)

Recognition Cues

• “Show there is a differentiable function $g:\mathbb R\to\mathbb R$ extending $f$” — construct and verify the extension at the patch point.
• “Is $f(x)$ differentiable at $x=a$? Prove your answer” — compute both one-sided derivatives.
• Function involves $x^2\sin(1/x)$ near $0$, or an absolute value of a trigonometric expression at a zero.

Solution Template

1. Identify the special point. Determine where differentiability might fail (patch point, zero of the inner function, etc.).
2. Handle $x\ne$ special point. Show differentiability away from the special point using the product/chain/sum rules.
3. One-sided derivatives at the special point. Compute $f'_-$ and $f'_+$ separately, either via the relevant branch formula or directly from the limit definition.
4. First-principles at the special point (if formulas diverge). Write $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ explicitly; apply the squeeze theorem or algebra.
5. Compare and conclude. If $f'_-=f'_+$, $f$ is differentiable (state the value); if not, $f$ is not differentiable (state the one-sided values).

Worked Example

2016 Paper 2, 2016-P2-Q1b (10 marks)

For $f(x)=x^2\sin\dfrac{1}{x}$, $x>0$, show that there is a differentiable function $g:\mathbb R\to\mathbb R$ extending $f$.

Step 1 — Define the extension. Set

$g(x)=\begin{cases}x^2\sin\dfrac{1}{x}, & x\ne 0,\\[2mm] 0, & x=0.\end{cases}$

On $(0,\infty)$, $g=f$. The choice $g(0)=0$ is forced by continuity: $|g(x)|\le x^2\to 0$.

Step 2 — Differentiability for $x\ne 0$. By the product and chain rules,

$g'(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x},\qquad x\ne 0.$

Step 3 — Differentiability at $x=0$ (first principles). Since the formula for $g'(x)$ oscillates as $x\to 0$, we must use the definition:

$g'(0)=\lim_{h\to0}\frac{g(h)-g(0)}{h}=\lim_{h\to0}\frac{h^2\sin(1/h)}{h}=\lim_{h\to0}h\sin\frac{1}{h}.$

Since $\bigl|h\sin(1/h)\bigr|\le|h|\to 0$, the squeeze theorem gives $g'(0)=0$.

Conclusion. $g$ is differentiable at every $x\ne 0$ (Step 2) and at $x=0$ (Step 3).

$\boxed{\;g(x)=x^2\sin(1/x)\ (x\ne0),\ g(0)=0,\ \text{is differentiable on }\mathbb R,\ g'(0)=0.\;}$

Remark. $g'$ is not continuous at $0$ (the term $-\cos(1/x)$ oscillates), so $g\notin C^1$. Differentiability does not require $g\in C^1$.

2019 Paper 1, 2019-P1-Q2a (15 marks)

Is $f(x)=|\cos x|+|\sin x|$ differentiable at $x=\pi/2$? Prove your answer.

Step 1 — Resolve $|\cdot|$ near $\pi/2$. Near $\pi/2$, $\sin x>0$ so $|\sin x|=\sin x$. But $\cos x$ changes sign at $\pi/2$:

$f(x)=\begin{cases}\cos x+\sin x, & x<\tfrac{\pi}{2},\\ -\cos x+\sin x, & x>\tfrac{\pi}{2}.\end{cases}$

Note $f(\pi/2)=0+1=1$.

Step 2 — Left-hand derivative.

$f'_-\!\left(\tfrac{\pi}{2}\right)=\lim_{x\to\frac{\pi}{2}^-}(-\sin x+\cos x)=-1+0=-1.$

Step 3 — Right-hand derivative.

$f'_+\!\left(\tfrac{\pi}{2}\right)=\lim_{x\to\frac{\pi}{2}^+}(\sin x+\cos x)=1+0=1.$

(Via difference quotient with $h\to 0^+$: $\dfrac{|\sin h|+\cos h-1}{h}\to 1$; with $h\to 0^-$: $\to -1$.)

Step 4 — Conclusion. $f'_-(\pi/2)=-1\ne 1=f'_+(\pi/2)$, so $f$ is not differentiable at $x=\pi/2$.

$\boxed{\;f\text{ is NOT differentiable at }x=\tfrac{\pi}{2};\quad f'_-=-1,\quad f'_+=+1.\;}$

Common Traps

• At $x=0$ for the extension: you cannot read $g'(0)$ from the formula $2x\sin(1/x)-\cos(1/x)$ — that formula has no limit at $0$. Always return to the first-principles definition.
• For $x^2\sin(1/x)$: the bound $|h\sin(1/h)|\le|h|$ works because of the $x^2$ factor; the analogous extension of $x\sin(1/x)$ by $g(0)=0$ is continuous but not differentiable at $0$.
• Only $|\cos x|$ causes a kink near $\pi/2$ — since $\sin x>0$ there, $|\sin x|$ is smooth. Students who differentiate $|\sin x|$ incorrectly waste time.
• The function $f(x)=|\cos x|+|\sin x|$ is continuous at $\pi/2$ (both branches give $1$). Continuity does not imply differentiability; you must check the one-sided derivatives.
• For the absolute-value type: get signs right. For $x>\pi/2$, $|\cos x|=-\cos x$, so its derivative is $+\sin x$, giving $f'_+=\sin(\pi/2)+\cos(\pi/2)=1$.

functional-equation-derivative (1 question(s); 2025)

Recognition Cues

• ”$f(x+y)=f(x)f(y)$ for all real $x,y$; $f'(0)$ is given; show $f'(x)=c\cdot f(x)$; find $f(x)$.”
• The functional equation is multiplicative (exponential type) or additive (linear type).
• “Hence find $f(x)$” signals: prove the ODE, then solve it by separation of variables.

Solution Template

1. Find $f(0)$. Set $x=y=0$ in the functional equation to determine $f(0)$.
2. Derive $f'(x)$ from first principles. Write $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$; use the functional equation to factor out $f(x)$.
3. Identify the remaining limit as $f'(0)$. The factor $\lim_{h\to0}\frac{f(h)-f(0)}{h}=f'(0)$.
4. State the ODE $f'(x)=f'(0)\cdot f(x)$.
5. Solve the ODE by separation: $\frac{f'}{f}=c$, so $f(x)=Ae^{cx}$; apply $f(0)=1$ to get $A=1$.

Worked Example

2025 Paper 1, 2025-P1-Q1d (10 marks)

Given $f(x+y)=f(x)f(y)$ for all real $x,y$; $f(x)\ne 0$; $f'(0)=2$. Show $f'(x)=2f(x)$ for all $x$, and find $f(x)$.

Step 1 — Find $f(0)$. Set $x=y=0$: $f(0)=f(0)^2$, so $f(0)(f(0)-1)=0$. Since $f\ne 0$ everywhere, $f(0)\ne 0$, hence

$f(0)=1.$

Step 2 — Derive $f'(x)$ from first principles.

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{f(x)f(h)-f(x)}{h}=f(x)\lim_{h\to0}\frac{f(h)-1}{h}.$

Step 3 — Identify the remaining limit.

$\lim_{h\to0}\frac{f(h)-1}{h}=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}=f'(0)=2.$

Therefore

$f'(x)=2f(x)\qquad\text{for all }x.$

Step 4 — Solve the ODE. Separate variables: $\dfrac{f'(x)}{f(x)}=2$, so $\dfrac{d}{dx}\ln|f(x)|=2$, giving $f(x)=Ae^{2x}$. Apply $f(0)=1$: $A=1$.

$\boxed{\;f'(x)=2f(x);\qquad f(x)=e^{2x}.\;}$

Common Traps

• First prove $f(0)=1$ before writing $f'(0)=\lim_{h\to0}\frac{f(h)-1}{h}$ — the $-1$ comes from $f(0)=1$.
• The functional equation factor step: $f(x+h)=f(x)f(h)$, not $f(x)+f(h)$. The multiplicative structure is the key.
• Don’t use the derivative formula without the definition; the problem asks you to show $f'(x)=2f(x)$, which requires the first-principles limit argument.
• After solving $f(x)=Ae^{2x}$, apply the initial condition $f(0)=1$ to pin down $A=1$. Leaving $A$ unresolved loses marks.

Marks-Aware Writing

10-mark questions (2016, 2025): For the extension question — define $g$ explicitly, handle $x\ne 0$ by rules (one line), then write out the first-principles computation at $0$ with the squeeze theorem step; box the conclusion. For the functional-equation question — the three steps (find $f(0)$, derive $f'(x)=2f(x)$ with the limit factored explicitly, solve the ODE) cover all marks. Omitting $f(0)=1$ or skipping the first-principles limit loses 3–4 marks.

15-mark question (2019): Removing $|\cdot|$ on both sides (Step 1) and computing $f'_-$, $f'_+$ via the branch formulas each carry 4 marks; the conclusion (not differentiable, with both values stated) carries 3 marks. A student who writes only the branch formulas without checking the two-sided derivatives earns at most 8 marks.

Practice Set

• 2021-P1-Q1c (10 m) — — Hint: compare one-sided derivatives using first principles; the function is likely piecewise near the test point.