Differentiability

1. The derivative as a limit. A function $f$ is differentiable at $x=a$ if the following limit exists and is finite:

where: $f'(a)$ is the derivative (instantaneous rate of change) at $a$, and $h$ is a small increment added to $a$. The quotient inside the limit is the slope of the chord joining $(a, f(a))$ to $(a+h, f(a+h))$; as $h\to0$ that chord slope settles down to the slope of the tangent line, provided it settles to the same value however $h$ approaches zero.

2. Splitting the limit in two directions. Because a two-sided limit exists only when both one-sided limits exist and agree, differentiability is checked via the left-hand derivative (LHD) and right-hand derivative (RHD):

where: $Lf'(a)$ lets $h$ approach $0$ through negative values (approaching $a$ from the left) and $Rf'(a)$ through positive values (approaching from the right). $f$ is differentiable at $a$ exactly when $Lf'(a) = Rf'(a)$ and both are finite; that common value is then $f'(a)$.

3. Differentiability is stronger than continuity. Every differentiable function is automatically continuous, but not every continuous function is differentiable. Symbolically,

and the converse is false. This one-directional arrow is worth memorising as a shape: differentiability sits nested strictly inside continuity, so a suspicious point (usually created by a modulus, a greatest-integer function, or a piecewise join) must be checked for continuity and for matching one-sided derivatives — passing one test says nothing about the other.

4. Reading it geometrically. A differentiable point has a unique, non-vertical tangent line — no sharp corner, no cusp, no vertical tangent. At $x=0$, $f(x)=|x|$ is continuous (both one-sided limits of the function equal $0$) but not differentiable, because the graph's left branch has slope $-1$ and its right branch has slope $+1$: two different tangent directions meeting at a corner, so no single tangent exists there. A cusp is a related but distinct failure — the graph again comes to a sharp point, but now both one-sided slopes race off to the same infinity (both $+\infty$ or both $-\infty$) rather than settling to two different finite values; a vertical tangent is different again, where the slope tends to $\pm\infty$ from both sides. This syllabus concentrates on the plain-corner case seen with modulus and piecewise functions, where the two one-sided derivatives are finite but simply unequal.

5. Why the order of checking matters. Since differentiability is strictly the stronger condition, it can never be verified in isolation — you always reach it through continuity, never around it. This is why nearly every exam question phrased as "test whether $f$ is differentiable at $x=a$" secretly expects two answers: first, is $f$ continuous at $a$ at all, and only then, do the one-sided derivatives agree. Skipping the continuity check does not usually produce a wrong numerical answer when the function happens to be continuous, but it is an incomplete argument, and examiners award marks for stating it explicitly.

Problem. Find the values of $a$ and $b$ so that $f(x) = x^2$ for $x \le 1$ and $f(x) = ax+b$ for $x>1$ is differentiable at $x=1$.

Step 1 — Continuity condition first. Since differentiability requires continuity, set the two one-sided limits, and $f(1)$, equal: $\lim_{x\to1^-}x^2 = 1$ and $\lim_{x\to1^+}(ax+b) = a+b$. Equating gives $a+b=1$, labelled equation (i).

Step 2 — Left-hand derivative at $x=1$. For $x\le1$, $f(x)=x^2$, so $f'(x)=2x$ on that branch, giving $Lf'(1) = 2(1) = 2$.

Step 3 — Right-hand derivative at $x=1$. For $x>1$, $f(x)=ax+b$, so $f'(x)=a$ everywhere on that branch, giving $Rf'(1) = a$.

Step 4 — Equate the derivatives. Differentiability demands $Lf'(1) = Rf'(1)$, so $a = 2$.

Step 5 — Solve for $b$. Substituting $a=2$ into equation (i): $2+b=1$, so $b=-1$.

Final answer. $\mathbf{a=2,\ b=-1}$, and at that join the function is both continuous and differentiable, with $f'(1)=2$.

Competitive stretch

Test $f(x)=|x-2|$ at $x=2$ using the same four-step method, purely to see a case that fails. Continuity holds ($\lim_{x\to2^-}|x-2| = 0 = \lim_{x\to2^+}|x-2|$). But $Lf'(2) = \lim_{h\to0^-}\frac{-h-0}{h} = -1$ while $Rf'(2) = \lim_{h\to0^+}\frac{h-0}{h}=1$; since $-1\ne1$, $f$ is continuous at $x=2$ yet not differentiable there — the vertex of the V-shaped graph.

• Assuming continuity automatically guarantees differentiability — always check both one-sided derivatives separately near a modulus, greatest-integer, or piecewise join.
• Computing only $Lf'(a)$ or only $Rf'(a)$ and stopping — a valid conclusion needs both, and they must be finite and equal.
• Substituting $h=0$ directly into the raw difference quotient instead of simplifying algebraically first, which usually produces a false $0/0$ instead of the correct limit.
• In piecewise problems, matching only the derivatives while forgetting the continuity condition — an $a,b$ pair that satisfies derivative-matching but not function-value-matching is not a valid final answer.

Ask the tutor: want to see why $f(x)=|x|\sin x$ actually is differentiable at $x=0$ despite involving $|x|$, unlike $|x|$ alone — work through the difference-quotient limit side by side with the $|x|$ case.