The chain rule computes the derivative of a composed function, f(g(x)), by multiplying an "outer" derivative by an "inner" one. Run that same idea backward and it becomes the single most useful skill in this course. It means looking at an integrand and recognizing "this is what a chain-rule derivative looks like," before any integration technique gets applied at all. That recognition, not the algebra that follows it, is what this section trains.
It's like recognizing a word inside a longer word without being told where the boundary is. The expression 2x\cos(x^2) doesn't announce that it's a chain-rule derivative. You have to spot the x^2 sitting inside \cos, and notice its derivative 2x standing right next to it. That's the same skill as spotting "in" inside "instant," once you know to look for it. Once that word-inside-a-word habit clicks, an integrand almost announces its own inner function: the same recognition, just running in the opposite direction from computing a derivative.
The chain rule computes the derivative of a composed function: one function fed into another, like \cos(x^2), where squaring happens first and cosine second. Written out, it says \frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x): an outer derivative times an inner one. Running that rule "in reverse" means starting from an expression that already has this shape and asking "what composed function produced THIS?" The tell is structural: an outer function applied to an inner one, multiplied by that inner function's own derivative sitting right next to it. 2x\cos(x^2) has this shape: \cos applied to x^2, times 2x, which is x^2's own derivative. Miss that structure and the next section's entire technique, u-substitution, has nothing to attach to. This recognition step is the actual prerequisite, not a footnote to it.
Formula
Chain Rule
Recalled from Calculus I — Differential & Integral Calculus · The chain rule — first taught there.
If y = f(u) and u = g(x), then \frac{dy}{dx} = f'(g(x)) \cdot g'(x).
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Where does the chain rule apply?
Set b — the derivative of \sin(bx) always carries an extra factor of b out front, from the chain rule's inner derivative.
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Worked example
Spotting the chain-rule shape before integrating
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1
Which of these is the derivative of a composed function, and what is the inner function? \quad 3x^2 e^{x^3} \quad vs. \quad 3x^2 + e^{x^3}.
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2
The first expression is a PRODUCT: 3x^2 next to e^{x^3}. The second is a SUM. Sums are never chain-rule derivatives, since the chain rule only ever produces a product of an outer and inner derivative.
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3
In the product, e^{x^3} is e^u with inner function u = x^3, whose derivative is u' = 3x^2. That's the factor sitting beside it.
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4
So 3x^2e^{x^3} is \dfrac{d}{dx}\left[e^{x^3}\right] by the chain rule. 3x^2 + e^{x^3} is not a chain-rule derivative of anything in one piece. It would need to be split and handled term by term. \boxed{\text{inner function: } u = x^3}
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Where this goes next
The very next section
u-substitution formalizes this recognition step into a mechanical procedure. Everything trained here is the diagnosis. The next section is the treatment.
Check your understanding (counts toward your progress)
For f(x) = 9\sin(4x), find f'(0).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
f'(x) = 9\cdot4\cos(4x), and \cos(0)=1, so f'(0) = 9\cdot4 = 36.
Correct answer
Also accepted:
Why
In the expression 4x^3e^{x^4}, what is the inner function?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
4x^3 is the OUTER derivative's matching factor, not the inner function itself.
e^x isn't present here at all — the exponent is x^4, not x.
This mixes both pieces together instead of naming the single inner function.
Why
e^{x^4} is e^u with u=x^4 — the inner function is x^4, whose derivative 4x^3 sits right beside it.
Identify what's raised as the exponent on e -- that's the inner function whose derivative should be sitting right next to it.
For f(x) = 4\sin(5x), find f'(0).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
f'(x) = 4\cdot5\cos(5x), and \cos(0)=1, so f'(0) = 4\cdot5 = 20.
Correct answer
Also accepted:
Why