Differentiability of functions

A continuous graph can still have a sharp corner — think of ∣x∣|x| at the origin, where the pen never lifts but the direction suddenly flips. Differentiability is the sharper question: does the graph have one well-defined, non-vertical tangent line at a point, with no kink and no vertical spike? This idea sits at the heart of the Continuity and Differentiability chapter of Class 12, right after continuity has been nailed down. The tool that answers the question is the first-principles derivative, built from a limit of a difference quotient, split into a left-hand and a right-hand version. When those two one-sided limits exist and agree, the function is differentiable there, and every later technique in the chapter — chain rule, implicit differentiation, logarithmic differentiation, parametric and second-order derivatives — is built on this single definition. This article walks through the definition, proves the key theorem linking differentiability to continuity, and works the classic ∣x∣|x| example end to end.

The core insight

A function is differentiable at x=ax=a exactly when its left-hand and right-hand derivatives both exist and are equal, and this differentiability is strictly stronger than mere continuity — every differentiable function is continuous, but a continuous function can still have a corner where no derivative exists.

The core idea

Differentiability from first principles. A function ff is differentiable at x=ax=a if the following limit exists:

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}

where: f′(a)f'(a) is the derivative of ff at aa (the instantaneous rate of change), hh is a small increment added to aa, and the fraction is the difference quotient — the slope of the chord joining (a,f(a))(a, f(a)) and (a+h,f(a+h))(a+h, f(a+h)). As h→0h\to0 that chord slope settles down to the slope of the tangent line, if the limit exists at all.

Splitting into one-sided derivatives. Because a two-sided limit exists only when both one-sided limits exist and agree, the same idea splits into a left-hand derivative (Lf') and a right-hand derivative (Rf'):

Lf′(a)=lim⁡h→0−f(a+h)−f(a)h,Rf′(a)=lim⁡h→0+f(a+h)−f(a)h\text{Lf}'(a) = \lim_{h\to 0^-} \frac{f(a+h)-f(a)}{h}, \qquad \text{Rf}'(a) = \lim_{h\to 0^+} \frac{f(a+h)-f(a)}{h}

where: h→0−h \to 0^- means hh approaches zero through negative values only, and h→0+h \to 0^+ means through positive values only. The function is differentiable at aa if and only if Lf′(a)\text{Lf}'(a) and Rf′(a)\text{Rf}'(a) both exist and are equal; that common value is then f′(a)f'(a).

The linking theorem. Differentiability and continuity are related but not identical: differentiability is the stronger condition. Every differentiable function is continuous, but a continuous function need not be differentiable — the graph can be unbroken yet still have a corner.

Geometric picture. Differentiability at a point means the graph has a single, well-defined, non-vertical tangent line there. A corner — where the tangent direction from the left disagrees with the tangent direction from the right — is continuous but not differentiable.

Figure
Two graphs side by side, a smooth curve with one tangent line at a point and a V shaped absolute value graph with two different tangent directions meeting at a sharp corner
One tangent versus two — differentiable versus merely continuous smooth curve: differentiable at P P single tangent line at P f(x) = |x|: corner at the origin two different tangent directions meet at the corner
Derivation AWhy differentiable functions must be continuous

Prove that if ff is differentiable at x=ax=a, meaning f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h} exists as a finite number, then ff must also be continuous at x=ax=a, meaning lim⁡h→0f(a+h)=f(a)\lim_{h\to 0} f(a+h) = f(a), and use this argument to explain why the converse implication can fail for a function such as ∣x∣|x|.

Step 1. Start from the assumption that f′(a)f'(a) exists as a finite limit:

f′(a)=lim⁡h→0f(a+h)−f(a)h,f′(a) finite.f'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}, \qquad f'(a) \text{ finite}.

This is the given hypothesis — the difference quotient settles down to a definite finite number as h→0h\to0, from both sides.

Step 2. Rewrite f(a+h)−f(a)f(a+h) - f(a) as the difference quotient multiplied back by hh, valid for every h≠0h \neq 0:

f(a+h)−f(a)=f(a+h)−f(a)h⋅h.f(a+h) - f(a) = \frac{f(a+h)-f(a)}{h} \cdot h.

This algebraic rearrangement is the whole trick — it turns the quantity we want to control, f(a+h)−f(a)f(a+h)-f(a), into a product of two pieces whose limits we already understand separately.

Step 3. Take the limit of both sides as h→0h \to 0, using the product rule for limits (the limit of a product is the product of the limits, when both exist):

lim⁡h→0[f(a+h)−f(a)]=(lim⁡h→0f(a+h)−f(a)h)⋅(lim⁡h→0h)=f′(a)⋅0=0.\lim_{h\to 0} \big[f(a+h) - f(a)\big] = \left(\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}\right) \cdot \left(\lim_{h\to 0} h\right) = f'(a) \cdot 0 = 0.

Since f′(a)f'(a) is a finite number by hypothesis, multiplying it by the vanishing factor h→0h\to0 forces the whole product to vanish.

Step 4. Conclude that lim⁡h→0f(a+h)=f(a)\lim_{h\to 0} f(a+h) = f(a), which is exactly the definition of continuity at x=ax=a. So differentiability at a point always forces continuity there — this is one-directional only. The converse fails precisely because continuity only guarantees the numerator f(a+h)−f(a)f(a+h)-f(a) shrinks to zero; it says nothing about the rate, and a sharp corner is a case where the numerator shrinks to zero along two different rates depending on the direction of approach, so the ratio in Step 1 never settles to a single value.

Worked example

Show that f(x)=∣x∣f(x) = |x| is continuous at x=0x=0 but not differentiable at x=0x=0, then find f′(3)f'(3) for f(x)=x2f(x)=x^2 using the same first-principles definition.

Step 1 (continuity check). f(0)=0f(0) = 0. As x→0−x\to 0^-, f(x)=−x→0f(x) = -x \to 0; as x→0+x\to 0^+, f(x)=x→0f(x) = x \to 0. Both one-sided limits equal 0=f(0)0 = f(0), so f(x)=∣x∣f(x)=|x| is continuous at x=0x=0.

Step 2 (left-hand derivative). For h<0h<0 near 00, f(0+h)=∣h∣=−hf(0+h) = |h| = -h since hh is negative, and f(0)=0f(0) = 0:

Lf′(0)=lim⁡h→0−−h−0h=lim⁡h→0−(−1)=−1\text{Lf}'(0) = \lim_{h\to 0^-} \frac{-h - 0}{h} = \lim_{h\to 0^-}(-1) = -1

Step 3 (right-hand derivative). For h>0h>0 near 00, f(0+h)=∣h∣=hf(0+h) = |h| = h:

Rf′(0)=lim⁡h→0+h−0h=lim⁡h→0+(1)=1\text{Rf}'(0) = \lim_{h\to 0^+} \frac{h - 0}{h} = \lim_{h\to 0^+}(1) = 1

Comparing, Lf′(0)=−1≠1=Rf′(0)\text{Lf}'(0) = -1 \neq 1 = \text{Rf}'(0), so f(x)=∣x∣f(x) = |x| is not differentiable at x=0x=0, even though it is continuous there — exactly the gap the theorem above predicts.

Competitive stretch

Step 4 (a genuinely differentiable case, for contrast). Take f(x)=x2f(x) = x^2 and find f′(3)f'(3) from first principles. Write f(3+h)=(3+h)2=9+6h+h2f(3+h) = (3+h)^2 = 9 + 6h + h^2 and f(3)=9f(3) = 9, so

f′(3)=lim⁡h→0(9+6h+h2)−9h=lim⁡h→06h+h2h=lim⁡h→0(6+h)=6.f'(3) = \lim_{h\to 0} \frac{(9+6h+h^2)-9}{h} = \lim_{h\to 0}\frac{6h+h^2}{h} = \lim_{h\to 0}(6+h) = 6.

Here the left-hand and right-hand limits both equal 66 automatically, since 6+h6+h is a single continuous expression with no absolute value or piecewise split — which is exactly why x2x^2 has no corner and is differentiable everywhere, unlike ∣x∣|x| at the origin.

Final answer: f(x)=∣x∣f(x)=|x| is continuous but not differentiable at x=0x=0 (Lf'(0) = -1, Rf'(0) = 1), while f′(3)=6f'(3) = 6 for f(x)=x2f(x)=x^2.

Common traps

  • Assuming continuity guarantees differentiability. This is the single most common conceptual error in the chapter — always keep ∣x∣|x| at 00 in mind as the standard counterexample.
  • Getting the sign of ∣h∣|h| wrong when computing one-sided derivatives: for h<0h<0, ∣h∣=−h|h| = -h, not hh; mixing this up flips the sign of the left-hand derivative.
  • Plugging in h=0h=0 too early. Substituting before simplifying the difference quotient gives the indeterminate form 00\frac{0}{0}, not the answer — the algebra in Step 2/3 of the worked example must be done first.
  • Confusing "the derivative exists" with "the derivative is zero." Differentiability is about the existence of the limit, not about its numerical size; a function can be differentiable everywhere with a large nonzero derivative.

Ask the tutor: Ask the AI tutor to sketch f(x)=∣x∣f(x)=|x| and f(x)=x2f(x)=x^2 side by side and trace how the left-hand and right-hand tangent lines behave at the origin for each.

Why does a sharp corner stop a function from being differentiable, even if it is continuous there?

Differentiability needs a single, well-defined tangent direction at the point. At a corner, the graph approaches from the left with one slope and from the right with a different slope, so the left-hand and right-hand derivatives disagree. Continuity only requires the graph to have no break or jump in height — it says nothing about matching slopes, which is exactly the extra condition differentiability adds.

Is every continuous function differentiable somewhere, or can it fail everywhere?

For Class 12 purposes, focus on functions with finitely many corners, like piecewise or absolute-value expressions, where differentiability fails only at those specific corner points and holds everywhere else. (Pathological continuous-but-nowhere-differentiable functions exist in higher mathematics, but they are well beyond this syllabus — you will only be tested on locating corners in simple piecewise graphs.)

Do I always need to check both continuity and differentiability separately, or does one check suffice?

Always check continuity as a quick first filter: if a function is not continuous at a point, it automatically cannot be differentiable there, so you can stop immediately (the theorem's contrapositive). But if it passes the continuity check, you still must separately compute Lf'(a) and Rf'(a) and confirm they agree — continuity alone never proves differentiability.

Before this

Where this leads