Differentiability

The derivative is a limit, and limits don't always exist. So some functions have points where there is no derivative: no well-defined slope, no tangent line you can write down. This lesson shows what those points look like, how differentiability relates to continuity, and how to test a piecewise function, which is a classic AP question.

What differentiable means

A function $f$ is differentiable at $x = a$ if $f'(a) = \displaystyle \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ exists as a finite number. A two-sided limit exists only when both one-sided limits exist and agree, so you can split the question in two:

• the left-hand derivative uses $h \to 0^-$ (secant lines from points to the left of $a$),
• the right-hand derivative uses $h \to 0^+$ (secant lines from points to the right).

The function is differentiable at $a$ exactly when both of these exist, are finite, and are equal. Visually, if you zoom in far enough on the graph near $a$, it looks like a straight, non-vertical line.

Three ways to fail

Corners

To the left of $x = 2$, the graph of $y = |x - 2|$ is the line $y = 2 - x$ with slope $-1$. To the right, it is $y = x - 2$ with slope $1$. The left-hand derivative is $-1$ and the right-hand derivative is $1$. They disagree, so $f'(2)$ does not exist. No matter how far you zoom in, the corner stays a corner.

Cusps and vertical tangents

Both of these graphs are continuous at $x = 0$, but the secant slopes blow up there. For $y = x^{1/3}$, the slopes approach $+\infty$ from both sides: the tangent line is vertical, and a vertical line has no slope. For $y = x^{2/3}$, the slopes approach $-\infty$ from the left and $+\infty$ from the right, making a sharp point called a cusp. In both cases $f'(0)$ does not exist.

Discontinuities

If $f$ has a hole, jump or vertical asymptote at $a$, it is not differentiable at $a$. This follows from an important theorem.

Here's why the theorem is true. For $x \ne a$,

$$f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a)$$

As $x \to a$, the fraction approaches $f'(a)$ and $(x - a)$ approaches $0$, so $f(x) - f(a) \to f'(a) \cdot 0 = 0$. That means $\displaystyle \lim_{x \to a} f(x) = f(a)$, which is the definition of continuity.

The contrapositive is often more useful: if $f$ is not continuous at $a$, it is not differentiable at $a$. Always check continuity first.

Testing a piecewise function

For a piecewise function built from polynomials (or other smooth pieces), the only suspicious point is where the rule changes. At that point:

1. Check continuity. The two pieces must meet: the left and right limits must both equal $f(a)$. If they don't, stop: $f$ is not differentiable.
2. Check that the slopes match. Differentiate each piece and evaluate both derivatives at $a$. If they agree, $f$ is differentiable at $a$.

Practice

Practice 1

Which statement is always true?

AIf $f$ is differentiable at $x = a$, then $f$ is continuous at $x = a$.BIf $f$ is continuous at $x = a$, then $f$ is differentiable at $x = a$.CIf $f$ is not differentiable at $x = a$, then $f$ is not continuous at $x = a$.DIf $f(a)$ is defined, then $f$ is differentiable at $x = a$.

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Practice 2

At which value of $x$ is $f(x) = |x + 3|$ not differentiable?

A$x = 3$B$x = 0$C$x = -3$D$f$ is differentiable everywhere.

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Practice 3

Let $f(x) = \begin{cases} x^2 + 1, & x < 1 \\ 2x, & x \ge 1 \end{cases}$. Which statement is true about $f$ at $x = 1$?

A$f$ is not continuous at $x = 1$.B$f$ is continuous and differentiable at $x = 1$.C$f$ is continuous but not differentiable at $x = 1$.D$f$ is differentiable but not continuous at $x = 1$.

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Practice 4

Let $f(x) = \begin{cases} 3x + 1, & x < 2 \\ x^2 + k, & x \ge 2 \end{cases}$. Find the value of $k$ that makes $f$ continuous at $x = 2$. (With that $k$, is $f$ differentiable at 2?)

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 5

Find the constants $a$ and $b$ that make $f(x) = \begin{cases} ax + b, & x < 1 \\ x^3, & x \ge 1 \end{cases}$ differentiable at $x = 1$. Enter your answer as $(a, b)$.

Your answer

Enter a point like (2, -3)

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Practice 6

The graph of $g(x) = x^{1/3}$ passes through the origin. Which best describes $g$ at $x = 0$?

A$g$ has a corner at $x = 0$.B$g$ is not continuous at $x = 0$.C$g$ is differentiable at $x = 0$ and $g'(0) = 0$.D$g$ is continuous at $x = 0$ and has a vertical tangent line there.

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Practice 7

Find all values of $x$ where $f(x) = |x^2 - 4|$ is not differentiable.

Your answer

Separate answers with commas, e.g. 2, -5

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Practice 8

Let $h(x) = \begin{cases} x^2 - 1, & x \le 0 \\ x + 1, & x > 0 \end{cases}$. Which statement is true?

A$h$ is not differentiable at $x = 0$ because $h$ is not continuous at $x = 0$.B$h$ is not differentiable at $x = 0$ because the one-sided slopes are $0$ and $1$, but $h$ is continuous.C$h$ is differentiable at $x = 0$.D$h$ is continuous everywhere but has a vertical tangent at $x = 0$.

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