1
A function of a function
In f(g(t)) the number g(t) is computed first and fed to f. With f(t) = t² and g(t) = 3t + 1, the composition is (3t + 1)². Which one is inside matters: g(f(t)) would be 3t² + 1, a different function.
Lesson · Rates of Change and the Derivative · The derivative as a function
The Chain Rule
When one function is fed into another, the rate of the whole is a chain of two rates. The outer function's derivative is evaluated at the inner function's value, and multiplied by the inner function's derivative. Without the second factor the answer is wrong by exactly that factor. For polynomials the rule is proved from the product rule and checked by expanding.
(f(g(t)))′ = f′(g(t))·g′(t)
outer rate, at the inner value
times the inner rate
(3t + 1)²: 2(3t + 1)·3 = 18t + 6
the 3 is the derivative of 3t + 1
(uⁿ)′ = n·uⁿ⁻¹·u′
The chain rule
Core ideas
What this lesson establishes
Three ideas: what a composition is, why its rate is a product of two rates, and why the outer derivative is read at the inner value.
2
Rates multiply along the chain
If t moves by a small step, the inner function moves by g′ times that step. The outer function then moves by f′ times what its input moved. The rate of the whole is the product of the two rates: the inner rate g′(t), and the outer rate f′ taken at the place the outer function is actually standing.
3
The outer derivative is read at g(t), not at t
f′(g(t)) means: differentiate f, then put g(t) into the result. For f(t) = t² the derivative is 2t, and at the inner value 3t + 1 it is 2(3t + 1). Replacing g(t) by t throws away the inside of the composition.
The material
Outer at the inner value, times the inner rate
A first composition worked two ways, the rule and its proof, a power of a function, and the shift that leaves the multiplier alone.
Take f(t) = t² and the inner function g(t) = 3t + 1, so that f(g(t)) = (3t + 1)². Multiplied out this is 9t² + 6t + 1, and the power rule gives the derivative at once: 18t + 6.
Now suppose the expansion is not wanted. The derivative of the outer function t² is 2t, and read at the inner value 3t + 1 it is 2(3t + 1) = 6t + 2. That is not 18t + 6; it is off by a factor of 3, and 3 is the derivative of the inner function 3t + 1. Multiplying the two, (6t + 2)·3, gives 18t + 6.
Theorem · The chain rule for polynomials
For polynomials f and g, (f(g(t)))′ = f′(g(t))·g′(t).
Write u for g(t). For f(t) = tⁿ the claim is (uⁿ)′ = n·uⁿ⁻¹·u′. It is true for n = 1, where it says u′ = u′.
If it holds for n, the product rule gives (uⁿ⁺¹)′ = (uⁿ·u)′ = n·uⁿ⁻¹·u′·u + uⁿ·u′ = (n + 1)·uⁿ·u′, which is the claim for n + 1.
So the claim passes from each power to the next: from n = 1 to 2, from 2 to 3, and on to every power. A polynomial f is a sum of constants times powers, and differentiating is done term by term, so the claim holds for every polynomial f.
□
The proof is the product rule applied again and again, which is why this lesson follows “The Product Rule”. It covers the cases the Subject uses. In words: the rate of the whole is the outer function's rate, taken where the outer function is standing, times the rate at which the inner function is carrying it.
Example · A power of a function
Let f(t) = t³ and g(t) = t² + 1. The outer derivative is 3t², read at the inner value it is 3(t² + 1)², and the inner derivative is 2t. The product is 6t·(t² + 1)², which expands to 6t⁵ + 12t³ + 6t.
Expanding first confirms it: (t² + 1)³ = t⁶ + 3t⁴ + 3t² + 1, and its derivative is 6t⁵ + 12t³ + 6t. The lab's equality tile reads equal.
A shift leaves the multiplier alone
With f(t) = t² and g(t) = t − 4 the composition is (t − 4)², and the rule gives 2(t − 4)·1 = 2t − 8. The inner derivative is 1, so the chain factor is invisible. That is why a shift of the input is easy to differentiate and a stretch is not: a shift moves the graph without changing how steep it is, and a stretch by 3 makes every slope 3 times as large.
The pattern for a stretch and shift together, f(k·t + c), is k·f′(k·t + c): the constant c never appears in the multiplier, and k always does.
Interactive
The quotient in h
Difference quotients of polynomials, by algebra
Method
Using the chain rule
Four moves. Leaving the inner function in place is the second, and forgetting the last is the common slip.
1
Name the outer and the inner function
In (t² + 1)³ the inner function is t² + 1 and the outer is the cube. The inner one is what you compute first.
2
Differentiate the outer, and leave the inner in place
Take the derivative of the outer function and put the inner function where its variable was: 3·(t² + 1)², not 3t².
3
Differentiate the inner function
The inner derivative of t² + 1 is 2t. It is a derivative, not the inner function repeated.
4
Multiply, then check
Multiply the two results. Expand f(g(t)) and differentiate it as a polynomial; the two answers must be the same polynomial.
Outer, inner and the factor between
Score 0 / 0Choose the answer you can justify.
t³ of t² + 1
The outer function is f(t) = t³ and the inner function is g(t) = t² + 1. Every line is exact.
f′(t) = 3t², so f′(g(t)) = 3(t² + 1)²
g′(t) = 2t
product: 3(t² + 1)²·2t = 6t(t² + 1)²
(t² + 1)² = t⁴ + 2t² + 1
6t·(t⁴ + 2t² + 1) = 6t⁵ + 12t³ + 6t
expand first: (t² + 1)³ = t⁶ + 3t⁴ + 3t² + 1
differentiate: 6t⁵ + 12t³ + 6t the same
Both routes give the same polynomial, and the lab's rule tile prints it for this pair. At t = 1 the slope is 6 + 12 + 6 = 24.
For a rehearsal, differentiate (2t + 5)³ by the rule: the outer derivative read at the inner value is 3(2t + 5)², and the inner derivative is 2.
Common mistakes
1
Differentiating the outside and forgetting the inner derivativeFor (3t + 1)² the outer derivative read at the inner value is 2(3t + 1) = 6t + 2. The derivative of the expanded 9t² + 6t + 1 is 18t + 6, three times as large. The missing 3 is the slope of the inner 3t + 1, and a stretch makes every slope as many times larger as the stretch.
2
Replacing the inner function by t in the outer derivativeFor (t² + 1)³ the outer derivative read at the inner value is 3(t² + 1)². Writing 3t² instead drops the inside. At t = 1 the true slope is 6·1·(1 + 1)² = 24, the lab's rule tile gives the same, and 3t²·2t at t = 1 is only 6.
3
Multiplying by the inner function instead of its derivativeThe factor to multiply by is g′, not g. For (3t + 1)², writing 2(3t + 1)·(3t + 1) gives a quadratic, 18t² + 12t + 2, when the true derivative 18t + 6 is a line.
Check your understanding
✓
Finish when you can differentiate a function of a function by f′(g(t))·g′(t) and confirm it by expanding.
You should be able to name the outer and inner functions, differentiate the outer one and read it at the inner value, multiply by the inner derivative, expand to verify, and say what goes wrong when the last factor is omitted.
Note: That completes the rules the course needs: the power, product and chain rules. The last two lessons use them to read how a graph bends, in “The Second Derivative”, and then to meet a function whose rate is a multiple of itself.