Difficulty: Intermediate | Prerequisites: Core chain rule techniques (see Part 1 notes), product and quotient rules, basic trig derivatives.
Once you can mechanically apply the chain rule to a formula, the next step is using it in contexts where the functions are given as tables, graphs, or real-world models. These problems test whether you understand what the chain rule is doing, not just whether you can execute it. They also introduce higher-order derivatives of composite functions and the idea that units multiply through the chain rule the same way the derivatives do. This is the material that separates students who memorised the formula from those who understand it.
The chain rule works even when functions are defined by tables or graphs rather than formulas: read the needed function values and slopes, then multiply. In applied problems, the chain rule connects rates measured in different units (litres per metre times metres per hour gives litres per hour). Higher-order chain rule derivatives require differentiating your first answer again, which often involves the product rule on top of the chain rule.
Higher-order derivative
The derivative of a derivative. The second derivative f''(x) measures how the rate of change itself is changing. Think of it as acceleration if f(x) is position: velocity is f'(x), and how velocity changes is f''(x).
Composite rate of change
When quantity A depends on B, and B depends on C, the rate of A with respect to C is (dA/dB) · (dB/dC). In simple terms, you are chaining two rates together, which is literally what the chain rule does.
Slope from a graph
The derivative of a linear segment on a graph is the slope of that segment, calculated as rise over run: (y2 - y1)/(x2 - x1). When the graph is piecewise linear, the derivative is constant on each segment.
Unit analysis in the chain rule
The units of dy/dx are (units of y) per (units of x). In a chain dy/dx = (dy/du)·(du/dx), the units of u cancel, leaving (units of y) per (units of x). This is a useful check: if your answer's units do not simplify correctly, something went wrong.
To find the second derivative of a composite function, first find y', then differentiate y' again. The second differentiation usually requires the product rule, because y' is typically a product of terms.
Example: y = 8tan(x/5). First derivative: y' = 8sec^2(x/5) · (1/5) = (8/5)sec^2(x/5). Second derivative: differentiate (8/5)sec^2(x/5) using the chain rule again. The derivative of sec^2(u) is 2sec(u)·sec(u)tan(u)·u' = 2sec^2(u)tan(u)·u'. So y'' = (8/5)·2sec^2(x/5)·tan(x/5)·(1/5) = (16/25)sec^2(x/5)tan(x/5).
When f, g, f', and g' are given in a table, you apply the chain rule formula directly by looking up the values you need.
Method: To find d/dx[f(g(x))] at x = a, compute f'(g(a)) · g'(a). This means: look up g(a) in the table, then look up f' at that value, and multiply by g'(a).
Example: Given a table with x = 1, 2, 3 and values for f, f', g, g':
x=1: f(1)=2, f'(1)=7, g(1)=2, g'(1)=6
x=2: f(2)=3, f'(2)=4, g(2)=8, g'(2)=7
x=3: f(3)=1, f'(3)=5, g(3)=1, g'(3)=2
Find d/dx[f(g(x))] at x = 1: Need f'(g(1))·g'(1). g(1) = 2, so f'(2) = 4, and g'(1) = 6. Answer: 4 · 6 = 24.
Find d/dx[f(f(x))] at x = 3: Need f'(f(3))·f'(3). f(3) = 1, so f'(1) = 7, and f'(3) = 5. Answer: 7 · 5 = 35.
When f and g are given as graphs (often piecewise linear), you read slopes from the graphs instead of computing derivatives algebraically.
Method: To find a'(x) where a(x) = f(g(x)), use a'(x) = f'(g(x)) · g'(x). Read g(x) from the graph of g. Then read the slope of f at that value. Multiply by the slope of g at x.
Example from the assignment graph: f is piecewise linear passing through roughly (0, 3), (2, 3), (4, 4). g is piecewise linear passing through roughly (0, 0), (2, 1), (4, 2).
For a'(3) where a(x) = f(g(x)): g(3) ≈ 1.5, slope of f near 1.5 = 0 (f is flat between 0 and 2), slope of g at 3 = 1/2. So a'(3) = 0 · (1/2) = 0.
For b'(2) where b(x) = g(f(x)): f(2) = 3, slope of g at 3 = 1/2, slope of f at 2 = 0 (f is flat). So b'(2) = (1/2) · 0 = 0.
The key skill is reading slopes accurately from the graph. On a linear segment, the slope is constant and equals rise/run.
These problems give you real-world functions and ask you to interpret the chain rule in terms of units.
Spherical tank example: V(h) = (1000π/3)h^2(12 - h) gives volume in litres as a function of water height h in metres. h(t) = 1 + sin(πt) gives height as a function of time in hours.
V'(h) = dV/dh = (1000π/3)(24h - 3h^2). At h = 1: V'(1) = (1000π/3)(24 - 3) = 7000π ≈ 21991 litres per metre. The unit is litres per metre because you are measuring how volume changes per unit of height.
h'(t) = dh/dt = πcos(πt). At t = 2: h'(2) = πcos(2π) = π metres per hour.
By the chain rule, dV/dt = (dV/dh)·(dh/dt) = 7000π · π = 7000π^2 ≈ 69087 litres per hour. The units check: (litres/metre)·(metres/hour) = litres/hour.
Widget production example: w(t) = 10t^(3/2) widgets after t days. c(w) = 100 + 10w + 10√w dollars for w widgets.
w'(t) = 15t^(1/2), in widgets per day. At t = 4: w'(4) = 15·2 = 30 widgets per day.
c'(w) = 10 + 10·(1/2)w^(-1/2) = 10 + 5/√w, in dollars per widget. At w = 80: c'(80) = 10 + 5/√80 ≈ 10 + 0.56 = 10.56 dollars per widget.
[c(w(t))]' = c'(w(t))·w'(t), in dollars per day. At t = 4 (where w = 80): [c(w(4))]' = 10.56 · 30 = 316.68 dollars per day.
Second derivative of a composite (general pattern)
If y = f(u) where u = g(x), then y' = f'(u)·u'. To get y'', differentiate y' using the product rule:
y'' = f''(u)·(u')^2 + f'(u)·u''
This formula is worth memorising. The first term comes from the chain rule on f'(u), and the second term from the fact that u' might itself depend on x.
Unit multiplication in the chain rule
(dA/dB) · (dB/dC) = dA/dC
The intermediate variable's units cancel, just like fractions. This is your sanity check on applied problems.
The applied chain rule is exactly how engineers and scientists track how fast a quantity changes when it depends on an intermediate variable. A factory manager who knows the cost per widget and the production rate per day can instantly compute the spending rate per day by multiplying the two, which is the chain rule in action. The same logic applies to medical dosing (drug concentration depends on absorption rate, which depends on time), climate modelling, and financial risk analysis.
Looking up the wrong row in a table problem. When computing f'(g(1)), students often evaluate f' at x = 1 instead of at x = g(1). The chain rule says to evaluate the outer derivative at the inner function's value, not at the original input.
Reading the wrong slope from a graph. Students sometimes read the y-value of f at a point when they need the slope of f at that point. On a piecewise linear graph, the slope is rise/run on the segment, not the height of the line.
Ignoring units in applied problems. If your answer to "rate of change of volume with respect to time" comes out in metres, something is wrong. The units of a derivative are always (units of output) per (units of input), and in a chain they multiply.
Forgetting the product rule when taking the second derivative. y' = f'(g(x))·g'(x) is a product of two functions of x. Its derivative is not just f''(g(x))·g''(x). You must use the product rule, which gives two terms.
⚠️ Table-based chain rule problems appear on nearly every MTH 132 exam. They are quick to grade and directly test whether you understand the chain rule beyond mechanical formula application.
⚠️ Applied chain rule problems with units are a common exam format. You will be asked to both compute a numerical answer and select the correct unit from a list. Get in the habit of checking units as you work.
⚠️ Second derivatives of composite functions are a step up in difficulty. Expect at least one on midterm or final exams. The most common error is omitting the product rule step.
⚠️ Graph-based problems test whether you can read slopes accurately. Practise computing rise/run from plotted points before the exam.
True or False: To find d/dx[f(g(x))] at x = 2 from a table, you evaluate f'(2)·g'(2). (False. You evaluate f'(g(2))·g'(2). The outer derivative is evaluated at g(2), not at 2.)
Fill in the blank: If dV/dh is in litres per metre and dh/dt is in metres per hour, then dV/dt is in ____ per ____. (Answer: litres per hour)
True or False: The second derivative of f(g(x)) is f''(g(x))·g''(x). (False. You need the product rule: f''(g(x))·[g'(x)]^2 + f'(g(x))·g''(x).)
Fill in the blank: On a piecewise linear graph, the derivative on each segment equals the ____ of that segment. (Answer: slope)
Q: Using the table (x=1: f=2, f'=7, g=2, g'=6; x=2: f=3, f'=4, g=8, g'=7; x=3: f=1, f'=5, g=1, g'=2), find d/dx[g(f(x))] at x = 2.
A: Need g'(f(2))·f'(2). f(2) = 3, so g'(3) = 2. f'(2) = 4. Answer: 2 · 4 = 8.
Q: A tank's volume is V(h) = 500h^2 litres, and the water height changes as h(t) = 3 + cos(t) metres. Find dV/dt at t = 0.
A: V'(h) = 1000h. h(0) = 3 + 1 = 4, so V'(4) = 4000 litres/metre. h'(t) = -sin(t), so h'(0) = 0 metres/hour. dV/dt = 4000 · 0 = 0 litres/hour.
Q: If y = 5tan(3x), find y''.
A: y' = 5sec^2(3x)·3 = 15sec^2(3x). For y'', differentiate: y'' = 15·2sec(3x)·sec(3x)tan(3x)·3 = 90sec^2(3x)tan(3x).
Q: Production is P(t) = 20t^2 items after t hours. Cost is C(P) = 50 + 3P + 2√P dollars. What is the spending rate dC/dt at t = 3?
A: P(3) = 180. P'(t) = 40t, so P'(3) = 120 items/hour. C'(P) = 3 + 1/√P. C'(180) = 3 + 1/√180 ≈ 3.075 dollars/item. dC/dt = 3.075 · 120 ≈ 368.94 dollars/hour.
Q: From a graph, f is a line from (0,1) to (4,5) and g is a line from (0,3) to (4,1). Find d/dx[f(g(x))] at x = 2.
A: Slope of f = (5-1)/(4-0) = 1. Slope of g = (1-3)/(4-0) = -1/2. g(2) = 3 + (-1/2)(2) = 2. f'(g(2)) = f'(2) = 1 (constant slope). g'(2) = -1/2. Answer: 1 · (-1/2) = -1/2.
Table and graph-based chain rule problems are direct preparation for related rates (Ch. 3.9), where you will connect rates of change of different quantities using the chain rule and information given numerically rather than as formulas. The unit-analysis skill practised here carries through to every applied calculus problem for the rest of the course. Higher-order derivatives of composites reappear in Taylor series and in the study of concavity and inflection points.
Chain rule table, chain rule graph, applied chain rule, units chain rule, second derivative chain rule, higher-order composite derivative, rate of change units, dV/dt, composite rate, piecewise linear derivative, slope from graph, f(g(x)) table, MSU MTH 132, Calculus I Section 2.5, related rates preview, widget production rate