Differentiability and the chain rule

What you will be able to do

• define the derivative at a point as a limit, and test differentiability with the left and right derivatives;
• explain why a differentiable function is continuous, and why the opposite is false;
• use the chain rule, also when there are three or more links.

The derivative as a limit

Pick a point $a$ on a graph. The line through $(a, f(a))$ and $(a + h, f(a + h))$ has slope $\dfrac{f(a + h) - f(a)}{h}$. As $h$ shrinks to zero, this line turns into the tangent. The derivative of $f$ at $a$ is

when the limit exists. Then $f$ is differentiable at $a$. The number $f'(a)$ is the slope of the tangent. British texts say gradient and American texts say slope. They mean the same.

Left and right derivatives

The number $h$ can be negative or positive. So there are two one-sided limits.

• The left derivative $f'(a^-)$ is the limit of the quotient as $h \to 0^-$.
• The right derivative $f'(a^+)$ is the limit as $h \to 0^+$.

The function is differentiable at $a$ exactly when both exist and are equal.

For $f(x) = |x|$ at $a = 0$ the quotient is $\dfrac{|h|}{h}$. It is $-1$ for negative $h$ and $1$ for positive $h$. The two derivatives differ, so $|x|$ is not differentiable at $0$. Its graph has a corner. For $f(x) = x^2$ the quotient is $\dfrac{h^2}{h} = h$, which tends to $0$ from both sides.

Three ways to fail

• A corner. The left and right derivatives exist but differ, as for $|x|$ at $0$.
• A vertical tangent. For $f(x) = x^{1/3}$ at $0$ the quotient is $h^{-2/3}$, which grows without bound. The tangent is vertical, so there is no slope.
• A break. If $f$ is not continuous at $a$, it cannot be differentiable there. The next section shows why.

Differentiable means continuous

If $f$ is differentiable at $a$, then $f$ is continuous at $a$. To see this, write

As $h \to 0$ the fraction tends to $f'(a)$ and the factor $h$ tends to $0$. The product tends to $0$, so $f(a + h) \to f(a)$.

The opposite is false. The function $|x|$ is continuous at $0$ but not differentiable there. So when you test differentiability, check continuity first. If it fails, you can stop.

The rules you already know

• The sum and difference: $(u \pm v)' = u' \pm v'$.
• The product: $(uv)' = u'v + uv'$.
• The quotient: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.
• The power: $(x^n)' = nx^{n-1}$.

These rules hold inside a piece where one formula applies. At a point where the formula changes, go back to the limit definition and use the left and right derivatives.

The chain rule

A function of a function has two links. Suppose $y = f(u)$ and $u = g(x)$. Then

In function notation this reads $\bigl(f(g(x))\bigr)' = f'(g(x))\,g'(x)$.

In words: differentiate the outside function, keep the inside as it is, and multiply by the derivative of the inside.

• For $y = (3x + 1)^5$ the outside is $u^5$ and the inside is $3x + 1$. So $y' = 5(3x + 1)^4 \times 3 = 15(3x + 1)^4$.
• For $y = \sin(x^2)$, $y' = \cos(x^2) \times 2x$.
• With three links, multiply three factors. For $y = \sin^3(2x) = (\sin 2x)^3$, $y' = 3\sin^2(2x) \times \cos(2x) \times 2 = 6\sin^2(2x)\cos(2x)$.

A worked example

Is $f(x) = x^3$ for $x \le 2$ and $f(x) = 12x - 16$ for $x > 2$ differentiable at $x = 2$?

1. Continuity first. $f(2) = 8$. The right piece gives $12(2) - 16 = 8$ at $x = 2$. So $f$ is continuous at $2$.
2. Left derivative. On the left $f'(x) = 3x^2$, so $f'(2^-) = 12$.
3. Right derivative. On the right $f'(x) = 12$, so $f'(2^+) = 12$.
4. Compare. The two derivatives are equal, so $f$ is differentiable at $2$ and $f'(2) = 12$.

Practise with new numbers

Try it in Python

The program estimates slopes from the left and from the right at $x = 0$ for three functions. Then it checks the chain rule on $\sin(x^2)$.

Common mistakes

• Forgetting the inner derivative. The derivative of $(3x + 1)^5$ is not $5(3x + 1)^4$. You must also multiply by $3$.
• Testing one side only. A corner shows up only when you compare the two sides.
• Believing a continuous function is differentiable. Continuity is needed, but it is not enough.
• Using the rules across a join. For a piecewise function, differentiate each piece, then compare the one-sided derivatives at the join.

Remember these

• $f'(a) = \displaystyle\lim_{h \to 0}\frac{f(a + h) - f(a)}{h}$. It exists when $f'(a^-) = f'(a^+)$.
• Differentiable implies continuous. Continuous does not imply differentiable.
• Corners, vertical tangents and breaks are the ways to fail.
• Chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du}\times\dfrac{du}{dx}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the left and right derivatives of $f(x) = |x - 2|$ at $x = 2$. Is $f$ differentiable there?

2. Differentiate $y = (5x - 3)^6$.

3. Differentiate $y = \sin(x^3)$.

4. True or false: if $f$ is continuous at $a$, then $f$ is differentiable at $a$.

5. Is $f(x) = x^{1/3}$ differentiable at $x = 0$?

References

• OpenStax, Calculus Volume 1, Chapter 3, sections on the derivative as a function and on the chain rule.
• NCERT, Mathematics Part I for Class XII, Chapter 5, section on differentiability.