Differentiability of Functions

Differentiability asks whether a function has one finite, well-defined derivative at a point. Continuity is necessary, but corners, cusps and vertical tangents show that continuity alone is not enough.

On this page (10 sections)

What differentiability means

A function is differentiable at a point when its derivative exists there as a finite two-sided limit. Graphically, the curve has one well-defined tangent slope at that point.

Differentiability is local. A function can be differentiable on most of its domain and fail at isolated points.

f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}

Both sides must approach one finite real number. An infinite slope is useful geometric information, but it is not a finite derivative under this convention. The test concerns difference quotients based at the point, not simply whether a formula for $f'(x)$ exists elsewhere.

Differentiability implies continuity

If $f'(a)$ exists, then $f$ must be continuous at $a$ . This gives a fast diagnostic: a discontinuity automatically rules out differentiability.

The converse is false. A function may be continuous and still have no unique finite tangent slope.

For $h\ne0$ , factor the change in output as

f(a+h)-f(a)=\frac{f(a+h)-f(a)}{h}\,h

If the quotient tends to the finite number $f'(a)$ , multiplying by $h\to0$ makes the output change tend to zero. Thus $f(a+h)\to f(a)$ . This argument explains the implication; it does not establish its converse.

Corners and cusps

The absolute-value graph meets at the origin with left slope negative one and right slope positive one. — Both branches meet, but their finite one-sided slopes disagree.
Diagram explained
The left branch follows $y=-x$ and the right branch follows $y=x$ . Their common point is $(0,0)$ . Continuity holds there, but differentiability fails because the two defining one-sided difference quotients are −1 and 1.
Absolute value.svg by Qef, after Ævar Arnfjörð Bjarmason. CC BY-SA 3.0. Font substituted with DejaVu Sans to render minus signs correctly. Geometry and mathematical labels unchanged; rasterized on white.

A corner has two different finite one-sided slopes. A cusp has slopes that become unbounded with opposing directions. Both prevent an ordinary finite derivative.

A cusp is not a finite corner

For $c(x)=|x|^{2/3}$ at zero, the right-hand difference quotient tends to $+\infty$ and the left-hand quotient to $-\infty$ . The curve stays continuous, but the opposing unbounded slopes prevent a finite derivative. Writing $|x|^{2/3}$ keeps the example real on both sides.

For $|x|$ , by contrast, both one-sided quotients are finite: −1 on the left and 1 on the right. They still fail the test because they differ.

Reference values for the absolute-value graph

Exact plotted values for $y=|x|$
$x$	$\|x\|$
-3	3
-2	2
-1	1
0	0
1	1
2	2
3	3

\lim_{h\to0^+}\frac{|h|}{h}=1,\quad\lim_{h\to0^-}\frac{|h|}{h}=-1

Vertical tangents

At a vertical tangent, secant slopes grow without bound. In ordinary single-variable calculus, the derivative is not a finite real number there, even though the curve may still be continuous. The function $\sqrt[3]{x}$ is a standard example at x=0.

Use the real cube root. Its difference quotient at zero is $\sqrt[3]{h}/h=1/|h|^{2/3}$ for $h\ne0$ . It tends to $+\infty$ from both sides. This is a vertical tangent, unlike the opposing signs in the cusp example. Neither has an ordinary finite derivative at zero.

Making a piecewise function differentiable

Consider the piecewise function:

f(x)=\begin{cases}x^2,&x\le1,\\ax+b,&x>1.\end{cases}

First match the values

The left value is $f(1)=1$ . The right-hand limit is $a+b$ , so continuity requires $a+b=1$ . Matching slopes before this step could leave a jump.

Then match the one-sided quotients

\frac{(1+h)^2-1}{h}=2+h\longrightarrow2\quad(h\to0^-)

\frac{a(1+h)+b-1}{h}=a+\frac{a+b-1}{h}\quad(h>0)

Once $a+b=1$ , the right quotient is exactly $a$ . To agree with the left derivative, choose $a=2$ . Then $b=1-a=-1$ . The function is differentiable at 1 with derivative 2.

a=2,\qquad b=-1

a+b=1,\quad a=2\Rightarrow b=-1

Endpoints and one-sided derivatives

At an endpoint of a restricted domain, courses sometimes discuss a one-sided derivative. That is useful, but it is distinct from the ordinary two-sided derivative used at interior points. State which convention a problem expects.

A practical differentiability test

Testing differentiability

Testing differentiability
Step	Question
1	Is the function continuous at the point?
2	Do the left and right difference quotients approach finite values?
3	Are those values equal?

If all three conditions hold at an interior point, the derivative exists there.

A drawn curve cannot resolve every case. The derivative definition is the deciding test when the picture is unclear. Also, differentiability does not require the derivative function itself to be continuous. Those are distinct properties; introductory pictures should not substitute “looks smooth” for the finite-limit test.

Practice questions

Is $|x|$ differentiable at zero?
Can a function with a jump be differentiable at the jump?
For $f(x)=x^2$ when $x\le1$ and $f(x)=ax+b$ when $x>1$ , find parameters for differentiability at 1.

Check your answers

No. The left difference quotient tends to −1 and the right to 1. These finite limits disagree.
No. The function is not continuous there, and continuity is necessary for differentiability.
Continuity gives $a+b=1$ . The left derivative is 2 and the right derivative, after continuity is imposed, is $a$ . Thus $a=2$ and $b=-1$ .

Sources and further reading

These references support the definitions, theorem assumptions and standard conventions used in this article.

OpenStax Calculus Volume 1: The Derivative as a Function. Differentiability, derivative functions and non-differentiable points.