The Chain Rule

You can differentiate $\sin x$ and you can differentiate $x^3$, but $\sin(x^3)$ is neither of those: it is the sine of a quantity that is itself changing. Functions built by feeding one function into another are called compositions, such as $e^{-0.2t}$, $\sqrt{1+x^2}$ and $\ln(\cos x)$. The chain rule differentiates every one of them, and it rests on a single fact: when one quantity drives another, their rates of change multiply.

Outside and inside

A composition $h(x) = f(g(x))$ is computed in two stages. The inner function $g$ turns $x$ into a number $u = g(x)$, and then the outer function $f$ turns $u$ into $f(u)$. To find the layers of a formula, ask what you would do last if you evaluated it at $x = 3$ by hand. For $\sin(x^3)$ you would cube 3 first and take the sine last, so the outside is $\sin u$ and the inside is $u = x^3$.

The first two rows use the same two pieces in opposite orders. $\sin(x^3)$ cubes the input before the sine acts; $\sin^3 x$ takes the sine first and cubes the output.

A product is not a composition. In $x^3\sin x$, the two pieces are computed separately from $x$ and then multiplied; neither one is fed into the other, so that function needs the product rule (topic 2.8).

Why the rates multiply

Suppose $u$ changes 3 times as fast as $x$, and $y$ changes 4 times as fast as $u$. A small nudge to $x$ is magnified by 3 on its way to $u$, and that change in $u$ is magnified by 4 on its way to $y$, so $y$ changes $3 \cdot 4 = 12$ times as fast as $x$. Gears work the same way: if gear A turns gear B 3 times per turn and gear B turns gear C 4 times per turn, then C turns 12 times for each turn of A. With $y = f(u)$ and $u = g(x)$, Leibniz notation writes this as

Here is where it comes from. Give $x$ a small change $\Delta x$. That changes $u$ by some $\Delta u$, which in turn changes $y$ by some $\Delta y$. As long as $\Delta u \neq 0$, ordinary algebra gives

Now let $\Delta x \to 0$. The second fraction approaches $g'(x)$. Because $g$ is differentiable at $x$, it is continuous there (topic 2.4), so $\Delta u \to 0$ as well, and the first fraction approaches $f'(u)$, the derivative of $f$ at $u = g(x)$. If $\Delta u$ happens to equal 0 for arbitrarily small $\Delta x$, the first fraction is undefined; a more careful proof patches that case and gets the same rule. The $du$'s in Leibniz notation look like they cancel, which is a good memory aid, but the limit argument is the actual reason.

The outside derivative is evaluated at $g(x)$, not at $x$, because $f$ never receives $x$. Its input is $u = g(x)$, so its rate of change has to be measured at the value of $u$ it is actually given.

You can test the rule on a function you can also expand. For $h(x) = (3x+1)^2$, the outside is $u^2$ and the inside is $3x + 1$, so

Expanding first gives $h(x) = 9x^2 + 6x + 1$ and $h'(x) = 18x + 6$, the same answer. Leaving off the inside derivative gives $2(3x+1) = 6x + 2$, which is too small by exactly the factor of 3 at which the inside changes.

Every rule you know, with an inside

Each derivative rule from Unit 2 becomes a chain rule pattern when its input is a function $g(x)$ instead of $x$: differentiate the outside as usual, keep $g(x)$ as its input, and multiply by $g'(x)$.

The last row is the outside derivative $\frac{1}{u}$ evaluated at $u = g(x)$, times $g'(x)$; it needs $g(x) > 0$ because $\ln u$ is defined only for $u > 0$. For example, $\frac{d}{dx}\ln(x^2+4) = \frac{2x}{x^2+4}$.

Four patterns are worth recognizing on sight:

• A linear inside contributes its slope. $\frac{d}{dx}\sin(5x) = 5\cos(5x)$ and $\frac{d}{dx}e^{-x/2} = -\frac{1}{2}e^{-x/2}$. The same holds for a function you know only by name: $\frac{d}{dx}f(2x) = 2f'(2x)$.
• Roots and reciprocals are powers. Rewrite $\sqrt{1+x^2}$ as $(1+x^2)^{1/2}$. Then its derivative is $\frac{1}{2}(1+x^2)^{-1/2}\cdot 2x = \frac{x}{\sqrt{1+x^2}}$.
• Exponentials with other bases. Since $b = e^{\ln b}$, you can write $b^x = e^{x\ln b}$, an exponential whose inside $x\ln b$ has slope $\ln b$. So $\frac{d}{dx}b^x = b^x\ln b$, and for instance $\frac{d}{dx}2^x = 2^x\ln 2 \approx 0.693 \cdot 2^x$.
• A power of a trig function is a power first. $\sin^2 x$ has outside $u^2$, so its derivative is $2\sin x\cos x$. Compare $\sin(x^2)$, whose derivative is $\cos(x^2)\cdot 2x$.

Three or more layers, and other rules mixed in

With three layers, the chain rule applies at each link, so the derivative is a product of three factors. Work from the outside in: differentiate the outermost layer while leaving everything inside it untouched, multiply by the derivative of what was inside, and repeat until you reach $x$. For $y = \ln(\cos 4x)$, on an interval where $\cos 4x > 0$, the layers are $\ln u$, then $u = \cos v$, then $v = 4x$:

When products, quotients and compositions are mixed, the operation you would do last decides which rule you use first. In $x^2e^{-3x}$ the last step is multiplying $x^2$ by $e^{-3x}$, so the product rule comes first, and the chain rule shows up inside it when you differentiate $e^{-3x}$:

Compositions given by tables and graphs

The exam can give $f$ and $g$ only through a table of values or a graph. The rule does not change; you look up each number instead of computing it. For $h(x) = f(g(x))$ at $x = a$:

1. Find the inside value $g(a)$.
2. Look up the outside derivative at that value, $f'(g(a))$.
3. Look up $g'(a)$ and multiply.

The two lookups come from different rows of the table, row $a$ for $g'$ and row $g(a)$ for $f'$, and reading $f'$ from row $a$ instead gives a wrong number.

On a graph, the derivative at a point is the slope there, so for a graph made of line segments, $f'$ at an input is the slope of the segment containing it. If $g(a)$ lands exactly on a corner of the graph of $f$, then $f$ is not differentiable at $g(a)$ and the chain rule's condition fails. When in addition $g'(a) \neq 0$, the composition really does have a corner at $a$: its slopes on the two sides of $a$ are the two segment slopes, each multiplied by $g'(a)$, and they disagree. Example 3 works through one.

Worked examples

Worked exampleExample 3: a composition built on a graph

The graph of the function $f$, which consists of three line segments, is shown below. Let $j(x) = [f(x)]^3$ and $h(x) = f(x^2+1)$.

(a) Find $j'(3)$.

(b) Find $h'\left(\sqrt{2}\right)$.

(c) Is $h$ differentiable at $x = 1$? Justify your answer.

Show solution

First read the slopes off the graph: $f' = 2$ on $(0, 2)$, $f' = -1$ on $(2, 4)$, and $f' = -\frac{3}{2}$ on $(4, 6)$.

(a) The outside of $j$ is $u^3$ and the inside is $f(x)$, so $j'(x) = 3[f(x)]^2 f'(x)$. The point $x = 3$ is the midpoint of the segment from $(2, 5)$ to $(4, 3)$, so $f(3) = 4$ and $f'(3) = -1$:

$$j'(3) = 3(4)^2(-1) = -48.$$

(b) Here the inside is $x^2 + 1$, so $h'(x) = f'(x^2+1)\cdot 2x$. At $x = \sqrt{2}$ the inside is $3$, which lies on the middle segment:

$$h'\left(\sqrt{2}\right) = f'(3)\cdot 2\sqrt{2} = -2\sqrt{2}.$$

(c) At $x = 1$ the inside is $2$, a corner of the graph of $f$, so the chain rule cannot be used there. Look at $h$ on each side instead. For $x$ slightly less than 1, $x^2 + 1$ is slightly less than 2, where $f(u) = 1 + 2u$; so $h(x) = 1 + 2(x^2 + 1) = 2x^2 + 3$, whose slope at $x = 1$ is $4$. For $x$ slightly greater than 1, $x^2 + 1$ is slightly more than 2, where $f(u) = 7 - u$; so $h(x) = 6 - x^2$, whose slope at $x = 1$ is $-2$. The slope of $h$ from the left of $x = 1$ is $4$ and from the right is $-2$. They are not equal, so $h$ is not differentiable at $x = 1$.

Where students lose points

On the exam

Questions that test the chain rule directly can give the composition as a formula, or give $f$ and $g$ only through a table or a graph so that you must look up $g(a)$, $f'(g(a))$ and $g'(a)$; they appear in multiple choice and as free-response parts. On no-calculator free-response parts, an unsimplified numeric answer such as $3(4)^2(-1)$ is acceptable; the arithmetic is not what is being tested. On calculator-active questions, the calculator can find a derivative's value at a point, but when a question asks for $f'(x)$ as an expression, you have to produce it by hand.