Why Diagnose the First Wrong Step?
A complicated derivative can contain several correct steps and still end with an incorrect answer. The most useful way to repair it is to find the first line that is not justified. Every later line may depend on that error, even if the later algebra is performed correctly.
Mixed expressions require particular care because their structure can contain several operations at once. A product in the numerator may need the product rule; a quotient may need the quotient rule; and a power or rational expression may contain a function that needs the chain rule. The task is not to apply every rule everywhere, but to identify which operation is outermost at each stage.
A Rule Check Before the Arithmetic
For differentiable functions \(u\) and \(v\), the product rule is \((uv)'=u'v+uv'\). Both terms are required. For differentiable \(u\) and \(v\), with \(v(x)\ne 0\), the quotient rule is \((u/v)'=(u'v-uv')/v^2\). The order and minus sign in the numerator matter. For differentiable functions \(g\) and \(h\), the chain rule is \((g(h(x)))'=g'(h(x))h'(x)\); the derivative of the inner function must be included.
These rules apply to different structures. The product rule handles two factors multiplied together. The quotient rule handles one function divided by another. The chain rule handles one function used as the input to another. A single expression may require more than one rule, applied at different levels.
One useful way to see why the quotient rule has its particular form is to write a quotient as a product with a reciprocal. If \(v(x)\ne 0\), then \(u/v=uv^{-1}\). Using the product rule and the chain rule for the reciprocal gives
This derivation also highlights a diagnostic check: the numerator must contain both \(u'v\) and \(-uv'\), while the denominator is \(v^2\). The condition \(v\ne0\) is necessary because the quotient is not defined where its denominator vanishes.
Worked Example: A Missing Inner Derivative
Worked Example: A Missing Inner Derivative
Consider \(f(x)=\frac{x^2+1}{(x-1)^2}\), defined for \(x\ne1\). A draft calculation reads:
Locate the first incorrect step and find the derivative.
Step 1: Identify the quotient and its denominator. Set \(u(x)=x^2+1\) and \(v(x)=(x-1)^2\). The quotient rule requires \(v'(x)\), not just the derivative of the outer square function with its input left unchanged.
Step 2: Differentiate the denominator using the chain rule. The outer function is \(z^2\), and the inner function is \(x-1\). Therefore
The draft's first incorrect step is using \(2\) in place of \(v'(x)=2(x-1)\). It has omitted the inner derivative's input factor \(x-1\).
Step 3: Apply the quotient rule with the correct derivatives. Since \(u'(x)=2x\),
Step 4: Simplify. Factor \(2(x-1)\) from the numerator:
Check. Rewrite \(f(x)=(x^2+1)(x-1)^{-2}\) and differentiate by the product and chain rules:
The independent calculation agrees with the quotient-rule result.
Worked Example: A Missing Product-Rule Term
Worked Example: A Missing Product-Rule Term
Let \(f(x)=\frac{(x^2+1)(2x-1)}{(x+2)^2}\), defined for \(x\ne-2\). A draft first sets \(N(x)=(x^2+1)(2x-1)\) and then claims \(N'(x)=2x(2x-1)\). What is the first error, and what is the correct derivative?
Step 1: Check the numerator. The numerator \(N\) is a product of \(x^2+1\) and \(2x-1\). The product rule requires a term for differentiating each factor. The draft's first incorrect step omits the term from differentiating \(2x-1\).
Step 2: Differentiate the product completely.
Also, for \(D(x)=(x+2)^2\), the chain rule gives \(D'(x)=2(x+2)\cdot1=2(x+2)\).
Step 3: Apply the quotient rule.
Here \(N(x)=2x^3-x^2+2x-1\), found by multiplying the two factors. Factor one \(x+2\) from the numerator and cancel it, which is allowed because \(x\ne-2\):
Check. At \(x=0\), the original numerator has value \((1)(-1)=-1\), its derivative has value \(2\), the denominator has value \(4\), and its derivative has value \(4\). The quotient rule gives
The simplified derivative also gives \(f'(0)=6/8=3/4\). This point check supports the calculation; the full rule-by-rule derivation establishes the formula for every \(x\ne-2\).
Worked Example: A Quotient Error Inside a Chain Rule
Worked Example: A Quotient Error Inside a Chain Rule
Let \(f(x)=\left(\frac{x^2+1}{x-1}\right)^2\), defined for \(x\ne1\). A draft sets \(u(x)=\frac{x^2+1}{x-1}\) and writes
Find the first incorrect step and correct the derivative.
Step 1: Check the outer operation. Since \(f=u^2\), the chain rule gives \(f'=2u\,u'\). The draft has this structure. Its first incorrect step is the formula it uses for \(u'\): a quotient derivative has the square of the denominator in its denominator, not the denominator to the first power.
Step 2: Differentiate \(u\) by the quotient rule. The numerator has derivative \(2x\), and the denominator has derivative \(1\). Thus
Step 3: Complete the chain rule.
Check. At \(x=0\), \(u(0)=-1\) and the quotient-rule expression gives \(u'(0)=(-1)/1=-1\). The chain rule therefore gives \(f'(0)=2(-1)(-1)=2\). Directly, write \(f(x)=\frac{(x^2+1)^2}{(x-1)^2}\). At \(x=0\), its numerator is \(1\), its numerator derivative is \(0\), its denominator is \(1\), and its denominator derivative is \(-2\). The quotient rule gives \(f'(0)=(0\cdot1-1\cdot(-2))/1^2=2\), in agreement.
A Reliable Diagnostic Sequence
When reading a mixed derivative calculation, inspect the structure before simplifying. Work from the outside operation inward, and check the derivative associated with each operation. A correct final-looking expression does not repair an unjustified earlier step: algebraic simplification preserves an expression, but it cannot turn a false derivative rule into a valid one.
Identify products, quotients, and compositions such as a power applied to a quotient.
For each line, ask whether the outermost operation calls for the product, quotient, or chain rule.
Look for both product-rule terms, the two signed quotient-rule terms, and the inner derivative in a composition.
Replace the first invalid line and derive the remaining steps anew, keeping the original domain restrictions.
Use a second representation, a separate application of the rules, or a carefully chosen point where the function is defined.
Common Pitfall: Correct Algebra After an Incorrect Derivative
A frequent mistake is to focus on whether the algebra after a derivative step is tidy. For instance, the expression in the first example may be simplified correctly after replacing the derivative of \((x-1)^2\) with \(2\). That simplification does not validate the result: the derivative should be \(2(x-1)\). The chain rule depends on the full inner function, not merely on the outer power.
Likewise, omitting one product-rule term can produce a plausible polynomial, and using \(v\) instead of \(v^2\) in a quotient-rule denominator can still lead to further algebra that looks orderly. Diagnose the rule before checking the algebra. Then preserve the domain: if the original denominator is zero at a point, canceling a factor in a simplified derivative does not make the original function defined there.
Check Your Understanding
For each question, identify the rule or condition that should be checked before simplifying.
- In differentiating \((x^2+3)(x-4)\), what two terms must appear in the product-rule result?
- A calculation differentiates \((3x+1)^4\) as \(4(3x+1)^3\). What factor is missing, and which rule supplies it?
- For \(u(x)/v(x)\), why must the calculation record the condition \(v(x)\ne0\)?
- In a mixed calculation, the quotient-rule denominator is correct, but the numerator omits the term involving \(u v'\). What is the first rule error?
- Why is checking a derivative at one value useful, but not enough by itself to prove a derivative formula for every point in its domain?
