Why Pointwise Chain-Rule Evaluation Matters
When a function is a composition, its derivative depends on two different inputs: the original input to the inner function, and the value produced by that inner function. At a specified point, keeping those inputs distinct is often more useful than first finding and simplifying a complete derivative formula.
If \(g\) is the inner function and \(f\) is the outer function, then the composition is \(f(g(x))\). The chain rule says to differentiate the outer function at the value \(g(x)\), then multiply by the derivative of the inner function at \(x\). At a particular input \(a\), this becomes a calculation in three parts: find \(g(a)\), find \(f'(g(a))\), and find \(g'(a)\).
The Pointwise Procedure
Use this order whenever you need the derivative of a composition at one specified input. Write down each value before multiplying; this makes it easier to catch an input mix-up.
Write the expression as \(f(g(x))\), so the roles of the two functions are explicit.
Compute \(g(a)\). This is the input at which the outer derivative must be evaluated.
Find \(f'(g(a))\) and \(g'(a)\), keeping their evaluation points visible.
Multiply the two derivative values. If useful, compare the result with the derivative formula for the composition.
The notation \(f'(g(a))\) is important: it means first compute \(g(a)\), then substitute that result into \(f'\). It does not mean evaluate \(f'\) at \(a\). The two inputs agree only in special cases.
Worked Examples
Worked Example: A Power of a Polynomial
Find the derivative of \(y=(3x^2-1)^4\) at \(x=1\).
Choose \(g(x)=3x^2-1\) and \(f(u)=u^4\), so \(y=f(g(x))\). First evaluate the inner function: $$ g(1)=3(1)^2-1=3-1=2. $$ The outer derivative is \(f'(u)=4u^3\). At the inner value \(u=2\), $$ f'(g(1))=f'(2)=4(2)^3=4\cdot 8=32. $$ The inner derivative is \(g'(x)=6x\), so $$ g'(1)=6(1)=6. $$ Multiply the values: $$ y'(1)=f'(g(1))g'(1)=32\cdot 6=192. $$ To check, differentiating the composition gives \(y'=4(3x^2-1)^3(6x)\). At \(x=1\), this is \(4(2)^3(6)=32\cdot6=192\), as required.
Worked Example: A Square Root Composition
Find the derivative of \(y=\sqrt{7-3x}\) at \(x=1\).
Set \(g(x)=7-3x\) and \(f(u)=\sqrt{u}\). The inner value is $$ g(1)=7-3(1)=4. $$ For \(u>0\), the outer derivative is \(f'(u)=\frac{1}{2\sqrt{u}}\). Since \(g(1)=4>0\), this derivative is defined at the required input, and $$ f'(g(1))=f'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4}. $$ The inner derivative is \(g'(x)=-3\), hence \(g'(1)=-3\). Therefore, $$ y'(1)=f'(g(1))g'(1)=\frac{1}{4}(-3)=-\frac{3}{4}. $$ For a check, the derivative formula is \(y'=\frac{-3}{2\sqrt{7-3x}}\). Substitution of \(x=1\) gives \(\frac{-3}{2\sqrt{4}}=-\frac{3}{4}\).
Worked Example: The Outer Derivative Is Not Evaluated at the Original Input
Let \(y=(2x+1)^3-2(2x+1)\). Find \(y'(2)\) by treating the expression as a composition.
Define \(g(x)=2x+1\) and \(f(u)=u^3-2u\), so \(y=f(g(x))\). At \(x=2\), the inner function has value $$ g(2)=2(2)+1=5. $$ Differentiate the outer function: \(f'(u)=3u^2-2\). Evaluate it at \(5\), not at \(2\): $$ f'(g(2))=f'(5)=3(5)^2-2=3(25)-2=75-2=73. $$ The inner derivative is \(g'(x)=2\), so \(g'(2)=2\). The chain rule gives $$ y'(2)=73\cdot2=146. $$ As a check, differentiating the written expression gives \(y'=3(2x+1)^2(2)-2(2)\). At \(x=2\), $$ y'(2)=6(5)^2-4=6(25)-4=150-4=146. $$
Worked Example: Evaluating Several Layers at One Point
Find the derivative of \(y=((x^2+1)^3+2)^2\) at \(x=1\).
Track the values from the inside outward. Let \(h(x)=x^2+1\), \(q(v)=v^3+2\), and \(r(w)=w^2\), so \(y=r(q(h(x)))\). At \(x=1\), $$ h(1)=1^2+1=2,\qquad q(h(1))=2^3+2=8+2=10. $$ The outer derivative is \(r'(w)=2w\), so its value at the output of \(q\) is \(r'(10)=2(10)=20\). The derivative of \(q\) is \(q'(v)=3v^2\), and it must be evaluated at \(v=h(1)=2\): $$ q'(2)=3(2)^2=3(4)=12. $$ Finally, \(h'(x)=2x\), so \(h'(1)=2\). Multiplying the three derivative values gives $$ y'(1)=20\cdot12\cdot2=240\cdot2=480. $$ For verification, differentiation by successive chain-rule applications gives $$ y'=2((x^2+1)^3+2)\cdot 3(x^2+1)^2\cdot 2x. $$ At \(x=1\), this equals \(2(10)\cdot3(2)^2\cdot2=20\cdot12\cdot2=480\).
Why the Evaluation Rule Works
The chain rule describes how small changes pass through a composition. A small change in \(x\) first changes \(g(x)\); that change in the inner output then changes \(f\). The two local rates of change multiply. The following argument makes this precise at a single point.
This argument also explains why the inner value matters: the outer function receives \(g(x)\) as its input, so its local rate of change must be measured at \(g(a)\). The inner function’s local rate, by contrast, is measured at the original input \(a\).
A Common Mistake: Using the Wrong Input
For \(y=(x^2+3)^4\) at \(x=1\), an incorrect calculation might use \(x=1\) directly in the outer derivative and write $$ 4(1)^3\cdot 2(1)=8. $$ This treats the outer function as though its input were \(x\). In fact, its input is \(g(x)=x^2+3\). At \(x=1\), that input is \(g(1)=1^2+3=4\). The correct evaluation is $$ f'(g(1))g'(1)=4(4)^3\cdot2(1)=4(64)\cdot2=512. $$ Checking by differentiation gives \(y'=4(x^2+3)^3(2x)\), and \(y'(1)=4(4)^3(2)=512\). The correction is to evaluate the inner function first, before substituting into the outer derivative.
When This Method Is Useful
Pointwise evaluation is especially efficient when a problem asks only for the derivative at one input. You can avoid expanding a composition or simplifying a full derivative, while still keeping every factor visible. It is also a reliable check on a derivative formula obtained by another route.
The method requires differentiability at the relevant points: \(g\) must be differentiable at \(a\), and \(f\) must be differentiable at \(g(a)\). For expressions involving roots or other restricted functions, also check that the inner value lies in a region where the outer derivative exists. A real square-root function, for example, has derivative \(\frac{1}{2\sqrt{u}}\) only for \(u>0\); evaluating this formula at \(u=0\) is not valid.
Before finishing a calculation, ask three questions: Did I compute \(g(a)\)? Did I evaluate \(f'\) at that value? Did I evaluate \(g'\) at \(a\)? If all three answers are clear, the chain-rule calculation is usually well organized.
Check Your Understanding
For each question, keep the inner value and the two derivative values distinct.
- For \(y=(x^2+2)^3\), find \(y'(1)\). Identify the inner value before evaluating the outer derivative.
- For \(y=\sqrt{5x+4}\), find \(y'(0)\). What is the input to the outer derivative at this point?
- For \(y=(3x-2)^5\), find \(y'(2)\) using the pointwise chain-rule formula.
- State the differentiability conditions needed to use the formula for the derivative of \(f(g(x))\) at \(x=a\).
- Explain why evaluating \(f'\) at \(a\) rather than \(g(a)\) is generally an error.
