What Does It Mean for a Function to Be Differentiable at a Point?

A function is differentiable at a point if its derivative exists there as a finite, unique value. Graphically, the curve must have one well-defined tangent gradient at that point.

For an interior point $x=a$, differentiability means that the limit

$$f'(a)=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$$

exists. The quotient represents the gradient of a secant through ((a,f(a))) and $(a+h,f(a+h))$. As $h$ approaches zero from both sides, these gradients must approach the same finite value.

Equivalently, the left-hand derivative and right-hand derivative must both exist and be equal. This rules out corners, cusps, jumps, and infinite gradients. A derivative is not merely a tangent line drawn by eye; it is the value of this two-sided limit.

For example, let $f(x)=x^2$. From first principles,

$$f'(a)=\lim_{h \to 0}\frac{(a+h)^2-a^2}{h} =\lim_{h \to 0}(2a+h)=2a.$$

Therefore, $f$ is differentiable at every real value of $a$. In practice, compare one-sided difference quotients whenever the formula changes at a point. If they disagree, or either side does not approach a finite number, the derivative does not exist.

A key result is

$$\text{differentiable at }a \Rightarrow \text{continuous at }a.$$

The converse is false. The function $f(x)=\lvert x\rvert$ is continuous at zero but not differentiable there. This corrects the common misconception that continuity automatically guarantees differentiability.

Exam technique: This is IB Mathematics AA HL, AHL 5.12: First principles, higher derivatives, relevant to Papers 1 and 2. Direct questions will not ask you to test continuity and differentiability, but the concept supports differentiation from first principles. Show the difference quotient, simplify before substituting $h=0$, and then evaluate the limit.