Why the outer structure matters
Every example so far had a single composition as its outer structure, so one chain rule block finished the derivative. A product such as
\[f(x)=(x^2+1)^7(3x-7)^4\]
has a different outer structure: the last operation performed on the whole expression is multiplication, not composition. The product rule therefore supplies the template, and the chain rule is used inside each factor.
Step 1: classify the outer structure
Ask what the last operation on the whole expression is.
- If it is multiplication of two non-constant pieces, the outer structure is a product.
- If it is division, the outer structure is a quotient.
- If it is a single function applied to a non-trivial argument, the outer structure is a composition.
For \((x^2+1)^7(3x-7)^4\), the last operation is multiplication, so the product rule applies.
Step 2: split into composite factors
The product rule operates on two factors:
\[u=(x^2+1)^7, \qquad v=(3x-7)^4\]
Each factor is itself a composition, so each one will need its own chain rule block. Neither factor is a plain polynomial, so neither derivative can be written down without the chain rule.
Step 3: mark the blocks before differentiating
Write the product rule skeleton with empty slots:
\[f'(x)=\big[,\text{block for }u,\big]\cdot v+u\cdot\big[,\text{block for }v,\big]\]
Each bracketed slot is one chain rule block. The factor \(v\) is copied unchanged into the first term, and \(u\) is copied unchanged into the second term. This is the same discipline as before: write the structure first, fill in the blocks second.
What this page establishes
A product of composites is not one chain rule problem — it is a product rule problem containing two chain rule problems. The next step is to expand each block using the labeled-product layout and place it into its slot.