The chain rule
When one quantity depends on a second, and the second depends on a third, how fast does the first change when the third does?
Where you are 13 in Derivatives
Why this exists
The rules so far handle sums, constant multiples, powers, exponentials, products and quotients. They do not handle a function inside another, such as (x^3 - 4x)^5, e^{x^2} or \displaystyle \sqrt{x^2 + 16}. Each is a composition: an inner rule is applied first and an outer rule is applied to its output.
The obvious guess, differentiate the outer function and leave the inside alone, already fails on (3x + 1)^2. The guess gives 2(3x + 1). Expanding gives 9x^2 + 6x + 1, whose derivative is 18x + 6 = 2(3x + 1) \cdot 3. The missing factor 3 is the rate at which the inside, 3x + 1, changes. The chain rule says that this is always the repair: multiply by the rate of the inside.
The same thing happens with units. Tumelo is descending on a scuba dive. The water pressure on him depends on his depth, and his depth depends on time. If the pressure rises 10 kilopascals for every meter of depth and he sinks 4 meters per minute, the pressure rises 10 \times 4 = 40 kilopascals per minute. Rates in a chain multiply, the way unit conversions do.
Definition
The chain rule. If g is differentiable at x and f is differentiable at g(x), then the composition f \circ g is differentiable at x, and
In words: the derivative of the outer function, evaluated at the inside, times the derivative of the inside.
In Leibniz notation, write u = g(x) and y = f(u). Then
where \displaystyle \frac{dy}{du} is evaluated at u = g(x), so that the final answer is in terms of x.
Intuition
Think of three gears in a row. If the middle gear turns 3 times for each turn of the first, and the last gear turns 2 times for each turn of the middle one, then the last gear turns 2 \times 3 = 6 times for each turn of the first. A derivative is the same kind of ratio, taken at an instant: \displaystyle \frac{du}{dx} is how many units u moves per unit of x, and \displaystyle \frac{dy}{du} is how many units y moves per unit of u.
The Leibniz form looks as if the du cancels, and that is a good way to remember it and to check units: (kilopascals per meter) times (meters per minute) is kilopascals per minute. But \displaystyle \frac{dy}{du} is a limit, not a fraction, and the cancellation is a memory aid. The proof below is what makes it true.
One point carries all the weight. The gear ratio \displaystyle \frac{dy}{du} is the one at the current value of u. For y = u^2 the ratio is 2u, which depends on where u is. That is why the outer derivative is evaluated at g(x) and not at x.
Results
Chain rule. (f \circ g)'(x) = f'(g(x))\,g'(x), when g is differentiable at x and f at g(x).
Powers of a function. For a differentiable g and any exponent n for which the power rule holds at g(x),
Exponentials. For a differentiable g, a constant k and a base a > 0,
Sine and cosine of a function. Using \displaystyle \frac{d}{dx}[\sin x] = \cos x and \displaystyle \frac{d}{dx}[\cos x] = -\sin x from the trigonometric derivatives,
Three layers. If y = f(u), u = g(v) and v = h(x), each differentiable at the right point, then
The chain grows one factor per layer, peeling from the outside in.
Why it is true
Fix x, write u = g(x), and for h \ne 0 let \Delta u = g(x + h) - g(x), so that g(x + h) = u + \Delta u.
The quick argument, and its gap. Multiply and divide by \Delta u:
As h \to 0, \Delta u \to 0 because g is differentiable, hence continuous, at x. The first factor tends to f'(u) and the second to g'(x). The gap: this divides by \Delta u, and \Delta u can be 0 for values of h as close to 0 as we like. A constant g does it for every h; Exercise 24 gives a function that does it infinitely often near 0 without being constant. Wherever \Delta u = 0 the first factor is not defined, so the argument is incomplete.
The repair. Define, for k \ne 0,
and set E(0) = 0. Because f is differentiable at u, E(k) \to 0 as k \to 0, and E(0) = 0, so E is continuous at 0. For k \ne 0 the definition rearranges to f(u + k) - f(u) = \big(f'(u) + E(k)\big)k, and for k = 0 both sides are 0. So this identity holds for every k, with no division. Put k = \Delta u and divide by h, which is never 0:
As h \to 0: \Delta u \to 0, so E(\Delta u) \to E(0) = 0 by continuity of E at 0, whether or not \Delta u is ever 0; and \displaystyle \frac{\Delta u}{h} \to g'(x). The product law for limits gives (f \circ g)'(x) = f'(g(x))\,g'(x).
The three-layer rule follows by applying the two-layer rule twice: f \circ g \circ h = f \circ (g \circ h), and (g \circ h)' = g'(h(x))\,h'(x).
Worked examples
Day 1: powers of a function
Differentiate (a) y = (x^3 - 4x)^5; (b) \displaystyle g(x) = \sqrt{x^2 + 16}, and find its tangent line at x = 3; (c) k(x) = \dfrac{3}{(2x + 1)^4} without the quotient rule.
Solution(a) The inside is u = x^3 - 4x and the outside is u^5. The outer derivative is 5u^4, evaluated at the inside, and the inner derivative is 3x^2 - 4:
Expanding the fifth power first would give a polynomial of degree 15 to differentiate term by term.
(b) Write g(x) = (x^2 + 16)^{1/2}. The outer derivative of u^{1/2} is \displaystyle \frac{1}{2}u^{-1/2}, and the inner derivative is 2x:
At x = 3, \displaystyle g(3) = \sqrt{25} = 5 and \displaystyle g'(3) = \frac{3}{5}. The tangent line through (3, 5) is \displaystyle y = 5 + \frac{3}{5}(x - 3), that is, \displaystyle y = \frac{3}{5}x + \frac{16}{5}.
(c) Write k(x) = 3(2x + 1)^{-4}. The constant multiple rule keeps the 3; the outer derivative of u^{-4} is -4u^{-5}, and the inner derivative is 2:
Day 1: exponentials through the chain rule
(a) Show that \dfrac{d}{dx}\big[e^{kx}\big] = k\,e^{kx} for a constant k. (b) Find f'(x) for f(x) = e^{1 - x^2} and every point where the tangent is horizontal. (c) Use a = e^{\ln a} to recover \dfrac{d}{dx}\big[a^x\big] = a^x \ln a, then differentiate q(x) = 2^{3x + 1} and find q'(0).
Solution(a) With u = kx, y = e^u has \displaystyle \frac{dy}{du} = e^u, and \displaystyle \frac{du}{dx} = k. So \displaystyle \frac{dy}{dx} = e^{kx} \cdot k. The lesson on exponentials reached this by rewriting e^{kx} as (e^k)^x; the chain rule gets it in one line and does not need the exponent to be linear.
(b) The inside is 1 - x^2, with derivative -2x:
Since e^{1 - x^2} > 0 for every x, f'(x) = 0 only at x = 0. The one horizontal tangent is at (0, e).
(c) Since a = e^{\ln a}, a^x = e^{x \ln a}, an exponential with k = \ln a. By part (a), its derivative is \ln a \cdot e^{x \ln a} = a^x \ln a, the rule that lesson proved from limits alone. Now q(x) = 2^u with u = 3x + 1:
So q'(0) = 3\ln 2 \cdot 2 = 6 \ln 2 \approx 4.159.
Day 1: Tumelo’s descent
Tumelo’s depth t minutes into a dive is d(t) = 6t - 0.5t^2 meters for 0 \le t \le 6. In a simple model of the water, the pressure on him at depth d meters is p = 101 + 10d kilopascals: 101 from the air above the water and 10 more for each meter of water. (a) How fast is the pressure rising at t = 2? (b) When does it stop rising, and what is it then?
Solution(a) Pressure depends on depth and depth on time, so
Here \displaystyle \frac{dp}{dd} = 10 kilopascals per meter at every depth, and \displaystyle \frac{dd}{dt} = 6 - t, which is 4 meters per minute at t = 2. So \displaystyle \frac{dp}{dt} = 10 \times 4 = 40 kilopascals per minute. The units multiply as the Leibniz form says: (kilopascals per meter)(meters per minute).
As a check, compose first: p = 101 + 10(6t - 0.5t^2) = 101 + 60t - 5t^2, whose derivative 60 - 10t is 40 at t = 2.
(b) \displaystyle \frac{dp}{dt} = 10(6 - t) is 0 exactly when t = 6, the moment Tumelo stops sinking. His depth is d(6) = 36 - 18 = 18 meters, and the pressure is 101 + 180 = 281 kilopascals.
Day 2: values from a table
The functions f and g are differentiable, with these values.
| x | f(x) | f'(x) | g(x) | g'(x) |
|---|---|---|---|---|
| 1 | 3 | -2 | 2 | 4 |
| 2 | 3 | 5 | 3 | -1 |
| 3 | 2 | 7 | 1 | 6 |
Find (a) (f \circ g)'(1), (b) (g \circ f)'(2), (c) the derivative of g(x)^3 at x = 3.
Solution(a) (f \circ g)'(1) = f'(g(1))\,g'(1). First find the inside value: g(1) = 2. The outer derivative is read at 2, not at 1: f'(2) = 5. So (f \circ g)'(1) = 5 \cdot 4 = 20.
(b) (g \circ f)'(2) = g'(f(2))\,f'(2). Here f(2) = 3, so the outer derivative is g'(3) = 6, and (g \circ f)'(2) = 6 \cdot 5 = 30.
(c) By the power form, the derivative is 3g(x)^2 g'(x), and at x = 3 this is 3 \cdot 1^2 \cdot 6 = 18.
The rule takes a value of the inside function, then a derivative of the outside function at that value. A table has to supply both.
Day 2: slopes read from graphs
The graphs of g and f are made of straight pieces.
Let u(x) = f(g(x)) and v(x) = g(f(x)). Find (a) u'(1), (b) u'(4), (c) v'(5), if it exists.
SolutionThe pieces have slopes: g has slope -2 on (0, 2) and \displaystyle \frac{1}{2} on (2, 6); f has slope 1 on (0, 3) and -1 on (3, 6). So g has a corner at x = 2, and f a corner at x = 3.
(a) g(1) = 2 and g'(1) = -2. The value 2 lies inside the first piece of f, so f'(2) = 1, and u'(1) = f'(2)\,g'(1) = 1 \cdot (-2) = -2.
(b) g(4) = 1 and \displaystyle g'(4) = \frac{1}{2}; f'(1) = 1. So \displaystyle u'(4) = 1 \cdot \frac{1}{2} = \frac{1}{2}.
(c) f(5) = 2 and f'(5) = -1, but g has a corner at 2, so g'(2) does not exist and the chain rule does not apply. Look at each side instead. Just left of x = 5, f(x) is slightly above 2, where g has slope \displaystyle \frac{1}{2}, so v has slope \displaystyle \frac{1}{2} \cdot (-1) = -\frac{1}{2} there. Just right of 5, f(x) is slightly below 2, where g has slope -2, so v has slope (-2)(-1) = 2. The one-sided slopes differ, so v'(5) does not exist: v has a corner at x = 5.
Day 2: three layers, and a product
(a) Differentiate y = \sin^3(4x). (b) Find k'(x) and \displaystyle k'\big(\frac{\pi}{4}\big) for k(x) = e^{\cos 2x}. (c) Find where f(x) = x\,e^{-x^2/2} has a horizontal tangent.
Solution(a) \sin^3(4x) means (\sin 4x)^3. Peel from the outside: a cube, then a sine, then 4x.
(b) The layers are e^u, u = \cos v, v = 2x, with derivatives e^u, -\sin v and 2:
At \displaystyle x = \frac{\pi}{4}, \displaystyle 2x = \frac{\pi}{2}, so \sin 2x = 1 and \cos 2x = 0: \displaystyle k'\big(\frac{\pi}{4}\big) = -2 \cdot 1 \cdot e^0 = -2.
(c) The outermost operation is a product, so the product rule goes first, and the chain rule handles e^{-x^2/2}, whose inner derivative is -x:
The exponential is never 0, so f'(x) = 0 exactly when x = 1 or x = -1. Deciding which rule applies first in a layered expression is the subject of combining the rules.
Day 2: Lesedi’s lift bag
Lesedi, Tumelo’s dive partner, carries a flexible bag of air. In the model, its volume in liters at pressure p kilopascals is V = \dfrac{1212}{p}, which is 12 liters at the surface pressure of 101. She rises from 18 meters at a steady 3 meters per minute, so her depth is d(t) = 18 - 3t for 0 \le t \le 6, and p = 101 + 10d as in Example 3. How fast is the bag growing at t = 4, and at t = 6?
SolutionVolume depends on pressure, pressure on depth, depth on time: three layers, so
The three rates are \displaystyle \frac{dV}{dp} = -\frac{1212}{p^2}, \displaystyle \frac{dp}{dd} = 10 and \displaystyle \frac{dd}{dt} = -3. Their product is
The first factor must be evaluated at the current pressure. At t = 4, d = 6 and p = 161, so \displaystyle \frac{dV}{dt} = \frac{36360}{25921} \approx 1.403 liters per minute. At t = 6 she is at the surface, p = 101, and \displaystyle \frac{dV}{dt} = \frac{36360}{10201} \approx 3.564 liters per minute.
The two negative rates multiply to a positive one: pressure falls as she rises, and the bag grows as pressure falls. Her speed is the same throughout, yet the bag grows more than twice as fast at the surface as at 6 meters, because \displaystyle \frac{dV}{dp} is larger in size where p is smaller.
Boundaries
- Both hypotheses matter. The rule needs g differentiable at x and f differentiable at g(x). In Example 5(c), f was differentiable at 5 but g was not differentiable at f(5) = 2, and the composition had a corner. The hypotheses are enough, not necessary: \displaystyle \sqrt{u} is not differentiable at u = 0, yet \displaystyle \sqrt{x^4} = x^2 is differentiable at x = 0; when a hypothesis fails, go back to the definition.
- The outer derivative is evaluated at the inside. f'(g(x)), not f'(x). In a table problem this means looking up g(a) first.
- The cancellation is not the proof. \displaystyle \frac{dy}{du} is a limit, not a quotient of two numbers, and dividing by \Delta u fails when \Delta u = 0. The function E in the proof is what closes the gap.
- Domains carry over. \displaystyle \sqrt{4 - x^2} is differentiable only for -2 < x < 2: at x = \pm 2 the inside is 0, where the square root has no derivative.
Where this goes next
The chain rule is what lets implicit differentiation treat y as an unnamed function of x: the derivative of y^2 is \displaystyle 2y\,\frac{dy}{dx}. Applied to f(f^{-1}(x)) = x, it gives the derivative of an inverse function, and from there the derivative of \ln x and the logarithmic and trigonometric derivatives for any inside. Rates that pass through an intermediate quantity, as in Lesedi’s bag, are the whole of related rates. The rule returns in parametric curves, where \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, and in the fundamental theorem when the upper limit is itself a function.
Exercises
Day 1: powers of a function
- y = (5x - 2)^7
- f(x) = (x^2 + 3x)^4
- \displaystyle g(x) = \sqrt{4 - x^2}, for -2 < x < 2.
- h(x) = \dfrac{1}{(x^3 + 1)^2}, without the quotient rule.
- Find the tangent line to y = (x^2 - 3)^3 at x = 2.
Day 1: exponentials
- y = e^{-4x}
- Find every x where f'(x) = 0 for f(x) = e^{x^3 - 3x}.
- g(x) = 10^{x^2}
Day 1: rates in a chain
- Neo sinks at a steady 0.4 meters per second, and the pressure rises 10 kilopascals per meter of depth. How fast is the pressure on him rising, per second and per minute?
- Neo pilots the dive boat at 12 miles per hour. It uses 0.5 gallons of gasoline per mile, and gasoline costs 4.20 dollars per gallon. At what rate, in dollars per hour, is he spending money on fuel?
- In Neo’s model of a lake, the water temperature at depth d meters is T = 8 + 16e^{-0.1d} degrees Celsius, and his depth after t minutes is d = 1.5t. How fast is the temperature around him changing at t = 4?
Day 2: values from a table
The functions f and g are differentiable, with these values.
| x | f(x) | f'(x) | g(x) | g'(x) |
|---|---|---|---|---|
| 0 | 1 | 3 | 2 | -2 |
| 1 | 2 | -1 | 0 | 5 |
| 2 | 0 | 4 | 1 | 3 |
- Find (f \circ g)'(0).
- Find (g \circ f)'(1).
- Find the derivative of f(f(x)) at x = 1.
- Find the derivative of f(g(f(x))) at x = 2.
Day 2: slopes read from graphs
Use the graphs of f and g in Example 5.
- Find (g \circ f)'(4).
- Find the derivative of f(2x) at x = 2.
Day 2: layers, trigonometry and other rules
- y = \cos(x^3)
- f(x) = \sin(e^{2x})
- h(x) = x^2 \sin(3x)
- \displaystyle q(x) = \dfrac{x}{\sqrt{x^2 + 1}}; simplify the answer to a single power.
- The water depth at the dive boat’s mooring is \displaystyle D(t) = 15 + 2\sin\big(\frac{\pi t}{6}\big) meters, t hours after midnight. How fast is it changing at t = 4?
Reasoning
- Use the chain rule on g(x)^{-1} to show \dfrac{d}{dx}\left[\dfrac{1}{g(x)}\right] = -\dfrac{g'(x)}{g(x)^2} wherever g(x) \ne 0 and g is differentiable, and compare with the quotient rule.
- Let \displaystyle g(x) = x^2 \sin\big(\frac{1}{x}\big) for x \ne 0 and g(0) = 0. Show that \Delta u = g(h) - g(0) is 0 for values of h as close to 0 as we like, and explain why this breaks the quick argument in “Why it is true” at x = 0.
- A function f is even if f(-x) = f(x) for every x. Show that the derivative of a differentiable even function is odd, that is, f'(-x) = -f'(x).
Answers
- y' = 35(5x - 2)^6. 7(5x - 2)^6 \cdot 5.
- f'(x) = 4(x^2 + 3x)^3(2x + 3). Outer 4u^3, inner derivative 2x + 3.
- \displaystyle g'(x) = -\dfrac{x}{\sqrt{4 - x^2}}. \tfrac{1}{2}(4 - x^2)^{-1/2} \cdot (-2x).
- h'(x) = -\dfrac{6x^2}{(x^3 + 1)^3}, x \ne -1. h = (x^3 + 1)^{-2}, so h' = -2(x^3 + 1)^{-3} \cdot 3x^2.
- y = 12x - 23. y(2) = 1; y' = 3(x^2 - 3)^2 \cdot 2x is 3 \cdot 1 \cdot 4 = 12 at 2; y = 1 + 12(x - 2).
- y' = -4e^{-4x}. Inner derivative -4.
- x = 1 and x = -1. f'(x) = (3x^2 - 3)e^{x^3 - 3x}, and the exponential is never 0.
- g'(x) = 2x \ln 10 \cdot 10^{x^2}. \displaystyle 10^u \ln 10 \cdot \frac{du}{dx} with u = x^2.
- 4 kilopascals per second, which is 240 kilopascals per minute. 10 \times 0.4, then \times 60.
- 25.20 dollars per hour. (dollars per gallon)(gallons per mile)(miles per hour): 4.20 \times 0.5 \times 12.
- About -1.32 degrees Celsius per minute: falling. \displaystyle \frac{dT}{dd} = -1.6e^{-0.1d}, which at d = 6 is -1.6e^{-0.6} \approx -0.878; times \displaystyle \frac{dd}{dt} = 1.5.
- -8. g(0) = 2, so f'(2)\,g'(0) = 4 \cdot (-2).
- -3. f(1) = 2, so g'(2)\,f'(1) = 3 \cdot (-1).
- -4. f(1) = 2, so f'(2)\,f'(1) = 4 \cdot (-1).
- -32. f(2) = 0 and g(0) = 2, so f'(2)\,g'(0)\,f'(2) = 4 \cdot (-2) \cdot 4.
- \displaystyle -\frac{1}{2}. f(4) = 3 and f'(4) = -1; \displaystyle g'(3) = \frac{1}{2}.
- -2. The derivative is 2f'(2x), and at x = 2 it is 2f'(4) = 2(-1).
- y' = -3x^2\sin(x^3). -\sin(x^3) \cdot 3x^2.
- f'(x) = 2e^{2x}\cos(e^{2x}). Three layers: \cos(e^{2x}) \cdot e^{2x} \cdot 2.
- h'(x) = 2x\sin(3x) + 3x^2\cos(3x). Product rule, with \displaystyle \frac{d}{dx}[\sin 3x] = 3\cos 3x by the chain rule.
- q'(x) = (x^2 + 1)^{-3/2}. Quotient rule: \displaystyle \dfrac{\sqrt{x^2 + 1} - x \cdot \frac{x}{\sqrt{x^2 + 1}}}{x^2 + 1}; multiply top and bottom by \displaystyle \sqrt{x^2 + 1} to get \dfrac{1}{(x^2 + 1)^{3/2}}.
- Falling at \displaystyle \frac{\pi}{6} \approx 0.524 meters per hour. \displaystyle D'(t) = \frac{\pi}{3}\cos\big(\frac{\pi t}{6}\big), and \displaystyle \cos\frac{2\pi}{3} = -\frac{1}{2}.
- With u = g(x) and y = u^{-1}: \displaystyle \frac{dy}{dx} = -u^{-2}\,g'(x) = -\dfrac{g'(x)}{g(x)^2}. The quotient rule with numerator 1 gives \dfrac{0 \cdot g(x) - 1 \cdot g'(x)}{g(x)^2}, the same.
- For \displaystyle h = \frac{1}{n\pi} with n a positive integer, \displaystyle \sin\frac{1}{h} = \sin(n\pi) = 0, so g(h) = 0 = g(0) and \Delta u = 0. These h approach 0 as n grows. At each of them the factor \displaystyle \frac{f(u + \Delta u) - f(u)}{\Delta u} divides by 0, so the quick argument cannot take the limit through those h. The repaired proof never divides by \Delta u.
- Differentiate both sides of f(-x) = f(x). The left side is f of -x, so by the chain rule its derivative is f'(-x) \cdot (-1). So -f'(-x) = f'(x), which is f'(-x) = -f'(x).