What it is
A function f is differentiable at the point a if the difference quotient (f(a + h) − f(a))/h has a finite limit as h tends to 0. The limit is the derivative f′(a); see the definition of the derivative. On the graph it shows as a tangent at the point that is not vertical. f is differentiable on an interval if it is differentiable at every point of the interval.
The limit exists only if the quotient approaches the same number from the right (h > 0) and from the left (h < 0). If the two one-sided limits differ, there is no derivative.
Examples of differentiable functions:
- f(x) = x². At a = 3 the quotient is ((3 + h)² − 9)/h = 6 + h, which tends to 6. The function is differentiable at every point, with f′(x) = 2x.
- Polynomials, sin x, cos x and eˣ are differentiable for every real x.
- A function that changes formula. f(x) = x² for x ≤ 1 and f(x) = 2x − 1 for x > 1. At x = 1 both expressions are 1, and the slope is 2 from both sides: the derivative of x² is 2 there, and the derivative of 2x − 1 is 2. f is differentiable at 1 with f′(1) = 2.
Differentiability fails at a point in three common ways:
- A corner. f(x) = |x| at a = 0: the quotient is |h|/h, which is 1 from the right and −1 from the left. The one-sided limits differ, so there is no derivative.
- A vertical tangent. f(x) = x^(1/3), the cube root of x, at a = 0: the quotient is h^(1/3)/h = h^(−2/3), which is 100 at h = 0.001 and grows without bound as h tends to 0. The graph has a vertical tangent at 0, but its slope is not a number, so the function is not differentiable there.
- A jump. A function that is not continuous at a is not differentiable there. f(x) = x for x < 1 and f(x) = x + 1 for x ≥ 1 jumps at x = 1. From the left the quotient is (h − 1)/h, which is 11 at h = −0.1 and 101 at h = −0.01, and it grows without bound.
Why it matters
- Differentiable implies continuous. If f is differentiable at a, then f(a + h) − f(a) = h · (f(a + h) − f(a))/h, and the product tends to 0 · f′(a) = 0 as h tends to 0. So f(a + h) tends to f(a), and f is continuous at a. The converse fails: |x| is continuous at 0 but not differentiable there.
- The rules for derivatives apply where the function is differentiable. f(x) = |x| has f′(x) = 1 for x > 0 and f′(x) = −1 for x < 0, but no derivative at 0.
- Tangents and approximation. Where f is differentiable there is a tangent, and near the point the tangent approximates the function.
- Extreme points. A minimum or maximum need not have f′ = 0: f(x) = |x| has its smallest value, 0, at x = 0, where there is no derivative. When looking for stationary points and extremes, check both the points where f′ = 0 and the points where the function is not differentiable.
Plot the function in the graphing calculator: corners, jumps and vertical tangents show as points where the graph has no single slope. The derivative calculator shows the derivative where it exists.
Common misconceptions
- “Continuous means differentiable.” |x| is continuous at 0 but has a corner there. The implication runs the other way, from differentiable to continuous.
- “If I can differentiate the expression, the function is differentiable.” The power rule gives (1/3) · x^(−2/3) for f(x) = x^(1/3) when x ≠ 0, but no value at x = 0. A formula for the derivative says nothing about the points where it has no value.
- “A vertical tangent means an infinite derivative, so the derivative exists.” The derivative is a number. At a vertical tangent there is no such number, and the function is not differentiable.
- “At a corner the derivative is the average of the slopes.” For |x| at 0 the slopes are −1 and 1 and their average is 0, but there is no derivative: the quotient is 1 or −1 for every h ≠ 0 and does not tend to 0.
Common questions
How do I check differentiability where a function changes formula?
First check that the function is continuous at the join, then that the derivative is the same from the left and
from the right. f(x) = x² for x ≤ 2 and f(x) = 4x − 4 for x > 2: both expressions are 4 at x = 2 and both
derivatives are 4, so f is differentiable at 2. g(x) = x² for x ≤ 2 and g(x) = 3x − 2 for x > 2: both
expressions are 4 at x = 2, but the derivatives are 4 and 3. g is continuous but not differentiable at 2.
Is f(x) = |x| differentiable?
At every point except x = 0. For x > 0, f(x) = x with derivative 1, and for x < 0, f(x) = −x with derivative
−1. At x = 0 the graph has a corner, and there is no derivative.
What does it mean for a function to be differentiable on an interval?
That it is differentiable at every point of the interval. The graph has no corners, jumps or vertical tangents
there.