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Differentiating a composite, why there are two factors, and the factor everyone forgets.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to name the outer and inner functions of a composite before differentiating either, apply the chain rule with the outer derivative evaluated at the inner function and the inner derivative present as a second factor, handle three or more layers with one factor per layer, recognise composites hidden inside roots, reciprocals and trigonometric expressions, and read the rule as a statement about rates multiplying through a chain.
If $f$ is differentiable at $g(a)$ and $g$ is differentiable at $a$, then $$\bigl(f \circ g\bigr)'(a) = f'\bigl(g(a)\bigr) \cdot g'(a).$$ In Leibniz notation, with $y = f(u)$ and $u = g(x)$: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.$$
Two things are easy to get wrong and they are both in the first line.
$f'$ is evaluated at $g(a)$, not at $a$. The outer derivative is asked about the place the inner function actually reached.
The second factor exists. $g'(a)$ is the most-forgotten symbol in first-year calculus. It is invisible in the easiest examples — when $g(x) = x + c$, $g' = 1$ and leaving it out changes nothing — which is exactly how the habit of omitting it forms.
Why there are two factors. Read it as rates. If $y$ changes $3$ times as fast as $u$, and $u$ changes $5$ times as fast as $x$, then $y$ changes $15$ times as fast as $x$. Rates through a chain multiply, and that is the whole content.
More layers, more factors. $\bigl(f(g(h(x)))\bigr)' = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$: peel from the outside in, one factor per layer, each outer derivative evaluated at everything still inside it.
Another way: picture
Three gears in a train. The first turns at some rate; the second is geared to it at $3:1$; the third to the second at $5:1$. The last gear turns $15$ times for one turn of the first — the ratios multiply along the chain. Reverse a gear and the sign flips, which is what a negative inner derivative does.
Another way: steps
Half the uses of the rule are in expressions nobody labels as composites.
| Expression | Inner function | Often mistaken for |
|---|---|---|
| $\sin(3x)$ | $3x$ | $\cos(3x)$, missing the $3$ |
| $\sqrt{x^2 + 1}$ | $x^2 + 1$ | $\dfrac{1}{2\sqrt{x^2+1}}$, missing the $2x$ |
| $e^{-kt}$ | $-kt$ | $e^{-kt}$, missing the $-k$ |
| $\ln(5x)$ | $5x$ | $\dfrac{1}{5x}$, missing the $5$ — though here the answer is $\dfrac1x$ either way, by luck |
| $\dfrac{1}{(x+2)^3}$ | $x + 2$ | handled by the quotient rule, four times the work |
The last row is worth a habit: a power of a bracket is a composite, and rewriting $\dfrac{1}{(x+2)^3}$ as $(x+2)^{-3}$ turns a quotient-rule problem into a chain-rule one.
And the $\ln(5x)$ row is worth a warning: an example where the missing factor does not change the answer teaches the wrong lesson. $\ln(5x) = \ln 5 + \ln x$, so its derivative really is $1/x$; the chain rule gives $\dfrac{1}{5x} \cdot 5$, which is the same thing. Do not generalise from it.
Leaving off $g'(x)$. The defining error of the chain rule. The defence is to write the inner function down before differentiating: it is hard to forget the derivative of something already on the page.
Evaluating $f'$ at $a$ rather than at $g(a)$. For $f(u) = u^2$ and $g(x) = 3x$ at $a = 2$: the answer is $2 \cdot g(2) \cdot 3 = 36$, not $2 \cdot 2 \cdot 3 = 12$.
*Differentiating both layers and multiplying the functions.* $(f \circ g)'$ is not $f' \circ g'$; it is $(f' \circ g) \cdot g'$. The composition and the multiplication happen in different places.
Miscounting layers. $\sin^2(3x)$ has three: squaring, sine, and $3x$. Three factors: $2\sin(3x) \cdot \cos(3x) \cdot 3$.
Treating $\dfrac{dy}{dx}$ as a fraction because the rule looks like cancellation. It is a theorem that happens to be well served by the notation. The notation is not the proof.
$\bigl((x+1)^2\bigr)' = 2(x+1)$, and the chain rule's second factor is $1$, so leaving it out gives the right answer. The same for $\sin(x + 3)$, for $\sqrt{x - 2}$, for $e^{x+1}$. A learner can do a page of these correctly with a rule that is wrong.
The error only shows when the inner function has a derivative other than $1$ — and by then the habit is formed. So the discipline has to be established on the easy cases, where it costs nothing and buys nothing: always write the inner derivative down, even when it is $1$. It is one symbol, and it is the difference between knowing the rule and having got away with not knowing it.
$f(x) = \sqrt{x^2 + 9}$. Rewrite as $(x^2+9)^{1/2}$, so the outside is $u^{1/2}$ and the inside is $x^2 + 9$.
The rewrite is what shows the structure.
Outer derivative at the inside: $\tfrac12 (x^2+9)^{-1/2}$. Inner derivative: $2x$.
Two pieces, written separately.
$f'(x) = \tfrac12 (x^2+9)^{-1/2} \cdot 2x = \dfrac{x}{\sqrt{x^2+9}}$. The $2$s cancelling is a coincidence of this example and not a rule.
$f(x) = \cos^3(2x)$, which is $\bigl(\cos(2x)\bigr)^3$: cube on the outside, cosine in the middle, $2x$ inside.
Name all three before starting.
Outermost: $3\bigl(\cos(2x)\bigr)^2$. Middle: $-\sin(2x)$. Innermost: $2$.
One factor per layer.
$f'(x) = 3\cos^2(2x) \cdot (-\sin(2x)) \cdot 2 = -6\cos^2(2x)\sin(2x)$.
Three layers, three factors — count them and check.
Inside $u = 4x^3 - 1$, outside $u^5$.
Name them first.
Outer derivative at the inside: $5(4x^3-1)^4$. Inner derivative: $12x^2$.
$f'(x) = 60x^2(4x^3 - 1)^4$.
Match each composite expression to the derivative factor its inner layer contributes.
| Inner factor $2x$ | Inner factor 4 | Inner factor $3x^2$ | Inner factor $-1$ | |
|---|---|---|---|---|
| $(x^2+1)^7$ | ||||
| $\sin(4x)$ | ||||
| $e^{x^3}$ | ||||
| $\ln(5-x)$ |
Match each part of $f(x)=\sin((3x-1)^2)$ to its role in a chain-rule calculation.
| Contributes $\cos((3x-1)^2)$ | Contributes $2(3x-1)$ | Contributes $3$ | Accounts for every nested dependency on $x$ | |
|---|---|---|---|---|
| The outer operation $\sin(u)$ | ||||
| The middle operation $u=v^2$ | ||||
| The inner operation $v=3x-1$ | ||||
| Multiply all three derivative factors |
For $f(x) = (2x + 2)^2$, write $f'(x)$.
Answer:
For $f(x) = \cos(x^2)$, which is the outer function and which the inner?
For $f(x) = \sin(4x)$, the derivative is $f'(x) = 4\cos(4x)$. What is $f'(0.7)$, to four decimal places?
Answer:
A balloon's volume grows at $3$ cm³ per cm of radius, and its radius grows at $9$ cm per second. How fast is the volume growing, in cm³ per second?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Differentiate $f(x) = (2x - 5)^{5}$, leaving the bracket unexpanded.
Answer:
You can differentiate a composite of two or three layers and say where each factor came from. Say in your own words why the second factor is the one that gets forgotten. Next: the derivatives of the trigonometric functions.
8. Your turn: differentiate $f(x) = (4x^3 - 1)^5$, step 3
Two layers, two factors.