Why the Order of the Lookups Matters
A derivative table gives information at particular inputs. For a composite function, the outer function is not necessarily evaluated at the original input \(x=a\). It is evaluated at the output of the inner function, \(g(a)\). That output determines which row of the outer function’s derivative table is needed.
This makes table problems less about complicated algebra and more about following the inputs carefully. The reliable order is: find the inner output, look up the outer derivative there, and multiply by the inner derivative at the original input.
The Chain Rule with Table Entries
The chain rule states that if \(g\) is differentiable at \(a\), and \(f\) is differentiable at \(g(a)\), then the composition \(H(x)=f(g(x))\) is differentiable at \(a\), with
Each factor comes from a different location. The inner derivative \(g'(a)\) is read at the original input \(a\). The outer derivative \(f'(g(a))\) is read at the inner output. The order of these lookups is essential.
Function values and derivative values also have different roles. The entry \(g(a)\) tells us where to look in the table for \(f'\). The entry \(f(g(a))\) gives the value of the composition, but does not by itself determine its derivative.
| Quantity | What it tells you | Where to find it |
|---|---|---|
| \(g(a)\) | The input to the outer function | The value table for \(g\), at input \(a\) |
| \(g'(a)\) | The inner function’s rate of change | The derivative table for \(g\), at input \(a\) |
| \(f'(g(a))\) | The outer function’s rate of change at the inner output | The derivative table for \(f\), at input \(g(a)\) |
| \(f(g(a))\) | The value of the composition | The value table for \(f\), at input \(g(a)\) |
Worked Example: A Two-Function Composition
Worked Example: A Two-Function Composition
Suppose \(H(x)=f(g(x))\), and the tables give the following entries:
| Input \(x\) | \(g(x)\) | \(g'(x)\) |
|---|---|---|
| 1 | 2 | \(-3\) |
| Input \(u\) | \(f(u)\) | \(f'(u)\) |
|---|---|---|
| 2 | 7 | 4 |
Find \(H'(1)\). First use the \(g\) table at the original input \(1\):
Since \(g(1)=2\), the needed outer derivative is \(f'(2)\). The \(f\) table gives \(f'(2)=4\). Therefore,
The value of the composition is also available: \(H(1)=f(g(1))=f(2)=7\). This value is distinct from the derivative. As an arithmetic check, the two derivative factors are \(4\) and \(-3\), whose product is \(-12\), so the composite’s rate of change is negative.
Why This Lookup Procedure Works
The chain rule describes how local rates combine. Near \(x=a\), a small change in \(x\) produces a change in \(g(x)\) at a rate of \(g'(a)\). The outer function then responds to that change at a rate of \(f'(g(a))\), because the outer function is being evaluated near \(g(a)\). Multiplying those two rates gives the rate of change of the composite.
This also gives a direct derivation of the table procedure. Start with the chain rule at \(a\):
A table supplies \(g(a)\) and \(g'(a)\) at the row for \(a\). Once \(g(a)\) is known, that value becomes the input for the lookup of \(f'(g(a))\). Substituting the two table entries into the chain rule gives the answer. Thus the inner output must be located before the outer derivative can be selected.
Worked Example: Reversing the Order of the Functions
Worked Example: Reversing the Order of the Functions
Let \(K(x)=p(q(x))\). Use these table entries to find \(K'(4)\):
| Input \(x\) | \(q(x)\) | \(q'(x)\) |
|---|---|---|
| 4 | \(-1\) | 5 |
| Input \(u\) | \(p(u)\) | \(p'(u)\) |
|---|---|---|
| \(-1\) | 6 | \(-2\) |
At the original input, the inner function has value and derivative
The outer derivative must be found at \(u=-1\), not at \(u=4\). The table gives \(p'(-1)=-2\). Applying the chain rule,
As a separate value check, \(K(4)=p(q(4))=p(-1)=6\). This does not replace the derivative calculation: the derivative is \(-10\), while the function value is \(6\). The derivative factors multiply to \(-10\), confirming the sign and arithmetic.
Compositions with More Than Two Layers
A composition can have several layers. For example, \(R(x)=A(B(C(x)))\) applies \(C\) first, then \(B\), and finally \(A\). The derivative uses one factor from each layer. Each outer derivative is evaluated at the output produced by all the layers inside it.
Applying the chain rule first to \(A(B(u))\), and then to the resulting composition with \(C(x)\), gives
The formula shows the lookup path clearly: start with \(x\), find \(C(x)\), then find \(B(C(x))\). Use these outputs to locate the corresponding derivative entries for \(B'\) and \(A'\).
Worked Example: Three Nested Functions
Worked Example: Three Nested Functions
Suppose \(R(x)=A(B(C(x)))\), with the following entries:
| Input \(x\) | \(C(x)\) | \(C'(x)\) |
|---|---|---|
| 0 | 2 | \(-1\) |
| Input \(u\) | \(B(u)\) | \(B'(u)\) |
|---|---|---|
| 2 | 5 | 3 |
| Input \(v\) | \(A(v)\) | \(A'(v)\) |
|---|---|---|
| 5 | 7 | \(-4\) |
Find \(R'(0)\). First, the innermost table gives
Now use \(C(0)=2\) as the input to the \(B\) table:
The resulting value \(B(C(0))=5\) is the input for the \(A\) table, which gives \(A'(5)=-4\). Therefore,
The composition’s value can also be followed through the value columns: \(C(0)=2\), \(B(2)=5\), and \(A(5)=7\), so \(R(0)=7\). For the derivative check, the three rate factors multiply as \((-4)(3)=-12\), followed by \((-12)(-1)=12\), agreeing with the result.
A Common Mistake: Looking Up the Outer Derivative Too Soon
A frequent error is to use the original input \(a\) as the input to the outer derivative table. For instance, in the first example, \(H(x)=f(g(x))\) and \(g(1)=2\). The incorrect lookup would use \(f'(1)\), even though the outer function receives the value \(2\).
The error changes which table entry is used; it is not fixed by multiplying carefully afterward. Correct the calculation by tracing the inputs in order: \(1\) goes into \(g\), \(g(1)=2\) goes into \(f\), and the outer derivative must therefore be read at \(2\).
Another common mix-up is to read \(f(g(a))\) when the problem asks for \(f'(g(a))\). The first is a function value; the second is a derivative value. Check the table’s column heading before using an entry.
A Reliable Routine for Table Problems
Identify which function is applied first and which function is outermost.
Use the inner function’s value table at that input to find the next input.
Read each derivative at the input where its function is evaluated, not automatically at the original input.
Use the chain rule and keep signs and factors attached to the correct layer.
Confirm that each lookup used the correct input and recompute the product, including its sign.
Why This Skill Matters
On an AP Calculus question, a function may be defined only by a table, so there may be no formula to differentiate. The chain rule still applies: the table gives the needed values and local rates, and the composition determines how those entries fit together.
Careful input-tracking also helps prevent a subtle but important error. A table entry is meaningful only together with its input. Before multiplying, be able to state where every factor came from: for example, “this is \(g'(a)\)” and “this is \(f'(g(a))\).” If the outer derivative is being read at the inner output, rather than at the original input by habit, the calculation is on the right track.
Check Your Understanding
For each question, identify the required table entries before calculating.
- If \(H(x)=f(g(x))\), \(g(3)=4\), \(g'(3)=2\), and \(f'(4)=-5\), find \(H'(3)\).
- If \(K(x)=p(q(x))\), which input is used to look up the outer derivative \(p'\) at \(x=a\)?
- If \(g(a)=6\), \(g'(a)=-2\), and \(f'(6)=3\), find \((f\circ g)'(a)\). What additional table entry would give \((f\circ g)(a)\)?
- For \(R(x)=A(B(C(x)))\), list the order in which you must determine the function outputs at a specified input \(x=a\).
- Explain why using \(f'(a)\) in place of \(f'(g(a))\) can select the wrong table entry.
