AP Calculus AB/BC · Unit 3 · Topic 3.1
The Chain Rule
Differentiate a composition by multiplying the outer derivative by the inner derivative.
1. Topic Focus
Differentiate nested, implicit, inverse, and higher-order relationships by choosing procedures that match the function structure.
This topic: Differentiate a composition by multiplying the outer derivative by the inner derivative.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
For sin(x²), the derivative is 2x cos(x²).
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
The chain rule differentiates a composition: one function's output becomes another function's input. If
then
Differentiate the outer function at the unchanged inner input, then multiply by the derivative of that inner input.
| View | Chain-rule statement | What it tracks |
|---|---|---|
| Function notation | \((f\circ g)'(x)=f'(g(x))g'(x)\) | Outer response times inner rate |
| Leibniz notation | \(\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}\) | Rates passing through an intermediate variable |
| At one point | \(h'(a)=f'(g(a))g'(a)\) | Evaluate the outer derivative at the inner output |
Why multiplication is reasonable. If a small change in \(x\) changes \(u=g(x)\) at about \(du/dx\) units of \(u\) per unit of \(x\), and changing \(u\) changes \(y=f(u)\) at about \(dy/du\) units of \(y\) per unit of \(u\), then the intermediate units cancel:
Difference-quotient intuition. Near \(x=a\), let \(u=g(x)\) and \(u_0=g(a)\). The composite difference quotient can be separated conceptually into
As \(x\to a\), the two factors approach \(f'(g(a))\) and \(g'(a)\). A rigorous proof also handles nearby points where \(u=u_0\), but this factorization shows the source of the two rates.
Common outer patterns.
| Composite function | Derivative |
|---|---|
| \([u(x)]^n\) | \(n[u(x)]^{n-1}u'(x)\) |
| \(\sin(u(x))\) | \(\cos(u(x))u'(x)\) |
| \(\cos(u(x))\) | \(-\sin(u(x))u'(x)\) |
| \(e^{u(x)}\) | \(e^{u(x)}u'(x)\) |
| \(\ln(u(x))\) | \(u'(x)/u(x)\) |
Layer routine.
- Identify the outermost operation.
- Keep its input unchanged while differentiating the outer function.
- Multiply by the derivative of that input.
- Repeat from outside to inside until reaching \(x\).
- Check the original and derivative domains.
A composition with three nontrivial layers produces three derivative factors. Never replace the inner input by its derivative inside the outer function: the rule uses \(f'(g(x))\), not \(f'(g'(x))\).
Choose the top-level rule first. A product such as \(x^2e^{\sin x}\) needs the product rule at the top level and the chain rule only inside the exponential factor. A quotient of composite expressions similarly begins with the quotient rule. Parentheses and the expression tree determine the order.
Domain matters. The composition must be defined, the inner function must be differentiable, and the outer function must be differentiable at the inner output. For example, \(\sqrt{1+\sin x}\) is defined when \(1+\sin x\ge0\), but its displayed derivative has a zero denominator where \(\sin x=-1\); those points require separate differentiability analysis.
6. Detailed Worked Example and Error Check
Example 1: Power of an inner polynomial. For \(y=(3x^2+1)^5\),
The factor \(6x\) is essential; omitting it differentiates only the outer power.
Example 2: Three layers. In \(y=\sin((x^2+1)^3)\), the layers are sine, cube, and \(x^2+1\):
Example 3: Exponential of a trigonometric function.
The exponential remains unchanged while the derivative of its exponent supplies \(-\sin x\).
Example 4: Logarithm of an algebraic function.
Because \(x^2+4>0\) for every real \(x\), both the original function and derivative are defined everywhere.
Example 5: Use values from a table. Suppose \(h=f\circ g\), \(g(2)=3\), \(g'(2)=-4\), and \(f'(3)=5\). Then
The needed outer derivative value is \(f'(3)\), because 3 is the output of the inner function at 2.
Example 6: Product rule combined with a chain. For \(F(x)=x^2e^{\sin x}\),
The product rule creates two terms. The chain rule creates the factor \(\cos x\) only in the term where \(e^{\sin x}\) is differentiated.
Example 7: Chain rule and a tangent line. Let \(p(x)=\sqrt{2x+7}\). At \(x=1\), \(p(1)=3\), and
The tangent line is \(y-3=\frac13(x-1)\).
AP error check. Do not omit an inner derivative, put the inner derivative inside the outer formula, differentiate from inside to outside, confuse a product with a composition, or evaluate \(f'\) at \(a\) when the formula requires \(f'(g(a))\).
7. AP Reasoning Routine
Mark inner and outer functions, track every derivative factor, solve algebraically for the requested derivative, and verify the result's domain.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Identify the layers before differentiating.
(a) Differentiate \(y=(5x-2)^4\).
(b) Differentiate \(y=\cos(x^3)\).
(c) Differentiate \(y=e^{\sin x}\).
(d) Differentiate \(y=\ln(1+x^2)\) and state its real domain.
(e) Differentiate \(y=\sqrt{1+(2x-1)^4}\).
(f) If \(H=f\circ g\), \(g(1)=-2\), \(g'(1)=3\), and \(f'(-2)=-5\), find \(H'(1)\).
(g) Differentiate \(R(x)=(x+1)\sin(x^2)\), naming the top-level rule.
(h) Find the tangent line to \(p(x)=\sqrt{2x+7}\) at \(x=1\).
Check the solution
(a) \(y'=4(5x-2)^3(5)=20(5x-2)^3\).
(b) \(y'=-\sin(x^3)(3x^2)=-3x^2\sin(x^3)\).
(c) \(y'=e^{\sin x}\cos x\).
(d) \(y'=2x/(1+x^2)\). Since \(1+x^2>0\), the domain is all real numbers.
(e) \(y'=4(2x-1)^3/\sqrt{1+(2x-1)^4}\).
(f) \(H'(1)=f'(g(1))g'(1)=(-5)(3)=-15\).
(g) The top-level rule is the product rule: \(R'(x)=\sin(x^2)+2x(x+1)\cos(x^2)\).
(h) \(p(1)=3\) and \(p'(1)=1/3\), so \(y-3=\frac13(x-1)\).