Why Local Rates Multiply
A quantity can change in stages. Time may change a radius, and that changing radius may change a volume. To find how quickly the volume changes with time, we need both links in the process: how quickly the radius changes with time, and how quickly volume changes with radius.
The chain rule gives a precise way to combine those rates. Its central idea is simple: for a small change in the input, the inner quantity changes by approximately its inner rate times the input change. The outer quantity then changes by approximately its outer rate times that inner change. Multiplying the two rates gives the resulting rate.
Composition and the Chain Rule
A composite function describes a quantity that depends on an input through an intermediate quantity. If \(u=g(x)\) and \(y=f(u)\), then the composite function is \(y=f(g(x))\). The variable \(x\) first determines \(u\), and \(u\) then determines \(y\).
The factor \(g'(a)\) is the inner rate: it describes how quickly \(u\) changes as \(x\) changes. The factor \(f'(g(a))\) is the outer rate, evaluated at the intermediate value \(u=g(a)\): it describes how quickly \(y\) changes as \(u\) changes. The chain rule multiplies these rates.
Units make the order of the multiplication especially clear. If \(x\) is measured in seconds, \(u\) in meters, and \(y\) in square meters, then \(du/dx\) has units of meters per second and \(dy/du\) has units of square meters per meter. Their product has units of square meters per second, the units required for \(dy/dx\).
The intermediate unit cancels. This does not replace the mathematical conditions of the theorem, but it is a useful check on whether the rates have been combined in the right order.
Why the Chain Rule Works
The derivative describes the best local linear approximation to a function near a point. Near \(x=a\), a small input change \(h\) produces an inner change approximately equal to \(g'(a)h\). The outer function then turns that inner change into an output change approximately equal to \(f'(g(a))\) times the inner change.
For a small input change \(h\), differentiability of \(g\) gives \(g(a+h)-g(a)=g'(a)h+o(h)\), where \(o(h)\) denotes a quantity whose ratio to \(h\) approaches zero as \(h\) approaches zero.
Write \(k=g(a+h)-g(a)\). Differentiability of \(f\) at \(g(a)\) gives \(f(g(a)+k)-f(g(a))=f'(g(a))k+o(k)\).
Since \(k=g'(a)h+o(h)\), the inner change is bounded by a constant times \(|h|\) near \(h=0\). Thus the outer remainder \(o(k)\) is also \(o(h)\), and $$ f(g(a+h))-f(g(a)) = f'(g(a))g'(a)h+o(h). $$
Divide by \(h\), then let \(h\) approach zero. The remainder divided by \(h\) approaches zero, leaving \(f'(g(a))g'(a)\), as claimed.
This argument explains the multiplication: the outer function responds to the change in the intermediate quantity, and that intermediate change is itself controlled by the inner rate. Differentiability at the two relevant points is essential to the theorem.
Worked Examples
Worked Example: A Changing Length and Its Square
Suppose a square has side length \(s(t)=5t+2\) meters, where \(t\) is measured in seconds. Its area is \(A(s)=s^2\) square meters. Find the area’s instantaneous rate of change at \(t=1\).
Step 1: Find the side length at the specified time.
Step 2: Find the inner rate. Differentiating \(s(t)=5t+2\) gives \(ds/dt=5\) meters per second.
Step 3: Find the outer rate at \(s=7\). Since \(A(s)=s^2\), \(dA/ds=2s\). Therefore, at \(s=7\), \(dA/ds=14\) square meters per meter.
Step 4: Multiply the rates.
Check: Substitution gives \(A(t)=(5t+2)^2=25t^2+20t+4\). Differentiating gives \(dA/dt=50t+20\), so at \(t=1\), \(dA/dt=50+20=70\) square meters per second. The result agrees, and the units also reduce correctly: \((\text{m}^2/\text{m})(\text{m}/\text{s})=\text{m}^2/\text{s}\).
Worked Example: Radius Changing with Time
A sphere’s radius is increasing according to \(r(t)=2+0.1t\) centimeters, where \(t\) is in minutes. Its volume is \(V(r)=\frac{4}{3}\pi r^3\). Find the volume rate at \(t=10\) minutes.
Step 1: Evaluate the radius.
Step 2: Find the inner rate. Differentiating \(r(t)=2+0.1t\) gives \(dr/dt=0.1\) centimeters per minute.
Step 3: Find the outer rate at \(r=3\).
Step 4: Multiply.
Check: Substituting \(r(t)\) into the volume formula gives \(V(t)=\frac{4}{3}\pi(2+0.1t)^3\). Differentiating this expression gives \(dV/dt=4\pi(2+0.1t)^2(0.1)\). At \(t=10\), this is \(4\pi(3)^2(0.1)=3.6\pi\) cubic centimeters per minute, which matches the chain-rule result. The units check as \((\text{cm}^3/\text{cm})(\text{cm}/\text{min})=\text{cm}^3/\text{min}\).
Worked Example: A Decreasing Side Length
A square’s side length is \(s(t)=20-3t\) meters, with \(t\) in seconds, and its area is \(A=s^2\). Find the area’s rate of change at \(t=2\), and interpret its sign.
Step 1: Evaluate the side length.
Step 2: Find both local rates. The inner rate is \(ds/dt=-3\) meters per second. The outer rate is \(dA/ds=2s\), so at \(s=14\), it is \(28\) square meters per meter.
Step 3: Multiply.
The negative sign means the area is decreasing at that instant. Check: Substituting gives \(A(t)=(20-3t)^2=400-120t+9t^2\). Thus \(dA/dt=-120+18t\), and at \(t=2\), \(dA/dt=-120+36=-84\) square meters per second. The direct calculation confirms the chain-rule result.
A Common Mistake: Leaving Out the Inner Rate
A frequent error is to differentiate the outer relationship while treating the intermediate quantity as if it changed at one unit per unit of input. For example, let \(u=2x+1\) centimeters, where \(x\) is in minutes, and let \(y=u^3\) cubic centimeters. At \(x=1\), the intermediate value is \(u=2(1)+1=3\) centimeters.
The outer rate there is \(dy/du=3u^2=3(3)^2=27\) square centimeters. But the inner rate is \(du/dx=2\) centimeters per minute. The incorrect calculation \(dy/dx=27\) has omitted that inner rate; it also has the wrong units for a rate of \(y\) with respect to time.
Check: Since \(y=(2x+1)^3\), differentiating gives \(dy/dx=3(2x+1)^2(2)\). At \(x=1\), this is \(3(3)^2(2)=54\) cubic centimeters per minute. The inner rate matters even when the outer rate has been calculated correctly.
Using Units as a Reasoning Tool
When a quantity depends on another quantity that depends on time, write down the units of each rate before multiplying. For instance, if \(V\) is in cubic centimeters, \(r\) is in centimeters, and \(t\) is in minutes, then \(dV/dr\) has units of square centimeters and \(dr/dt\) has units of centimeters per minute. The product has units of cubic centimeters per minute, as required.
Units can reveal that a factor is missing or that rates have been reversed. They cannot prove a derivative calculation is correct: two incorrect steps might still produce plausible units. Use them alongside the theorem’s conditions and, when possible, a direct check.
Check Your Understanding
For each question, identify the relevant rates and explain how they combine.
- If \(u=g(x)\) and \(y=f(u)\), which point determines where the outer derivative \(f'\) is evaluated?
- A length is measured in meters, time in seconds, and area in square meters. What are the units of \(dA/ds\), \(ds/dt\), and their product?
- A quantity has outer rate \(12\) units of output per unit of intermediate quantity and inner rate \(-2\) intermediate units per second. What is the composite rate, including its sign and units?
- What differentiability conditions are required at \(a\) for the chain rule used here?
- Why does omitting an inner rate generally produce an incorrect composite rate?
