Why Simplify Before Differentiating?
A rational expression is a quotient of polynomials. Sometimes its numerator and denominator contain a common factor, so the quotient rule produces more algebra than is needed. Factoring and cancelling that common factor first can leave a much simpler expression to differentiate.
There is an important qualification: cancellation does not change which inputs were excluded from the original expression. A simplified formula may be defined at an input where the original quotient was not. The derivative of the original function is still defined only at inputs in its domain.
The useful practice is therefore to do two things in order: record the domain restrictions, then simplify and differentiate. Afterward, check that the derivative formula applies on the original domain.
Domain Restrictions Come First
A rational expression \(P(x)/Q(x)\), where \(P\) and \(Q\) are polynomials, is defined at the real inputs \(x\) for which \(Q(x)\ne 0\). Before cancelling anything, find the zeros of the original denominator. Those inputs are excluded from the domain.
If the numerator and denominator have a common factor, cancellation is valid at inputs where that factor is nonzero. The resulting formula agrees with the original quotient at every input in the original domain. However, the new formula can have a larger domain because the cancelled factor no longer appears in its denominator.
Worked Example: Cancelling a Linear Factor
Consider \(f(x)=\frac{x^2-9}{x-3}\). First find the original domain: the denominator is zero when \(x-3=0\), so \(x\ne 3\).
Factor the numerator and cancel only on that domain:
Thus \(f'(x)=1\) for every \(x\ne 3\), since the derivative of \(x+3\) is \(1\).
We can check by applying the quotient rule to the expression as written. For differentiable \(u\) and \(v\), with \(v(x)\ne 0\), the quotient rule is \(\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\). Here \(u=x^2-9\), \(u'=2x\), \(v=x-3\), and \(v'=1\). Therefore, for \(x\ne 3\),
Both methods give \(f'(x)=1\) on the original domain. At \(x=3\), neither the original function nor its derivative is defined. The simplified formula \(x+3\) has a value there, but that value belongs to an extension of the original function, not to the original function itself.
Why Cancellation Preserves the Derivative on the Domain
The reason simplification works is more fundamental than any particular factoring pattern. A function’s derivative at an input is determined by the function’s values sufficiently close to that input. If two formulas agree throughout the original domain, then at any point in that domain they agree in a neighborhood of that point, and their difference quotients agree as well.
To justify this result, use the definition of the derivative. Since \(f(x)=g(x)\) for all \(x\) in the interval, and \(f(a)=g(a)\), the difference quotients agree whenever \(h\) is sufficiently close to zero:
Taking the limit as \(h\) approaches zero gives the same limit on both sides. Therefore \(f'(a)=g'(a)\), whenever those derivatives exist.
For a cancellation, consider \(f(x)=\frac{(x-c)A(x)}{(x-c)B(x)}\), where \(A\) and \(B\) are polynomials. The original quotient requires \((x-c)B(x)\ne 0\), so in particular \(x\ne c\) and \(B(x)\ne 0\). At every such input, cancellation gives \(f(x)=A(x)/B(x)\). For any point \(a\) in the original domain, the denominator remains nonzero in some interval around \(a\). The two formulas agree throughout that interval, so their derivatives at \(a\) agree.
This argument does not claim that the functions agree at \(x=c\), because the original function is not defined there. It establishes equality, and hence equality of derivatives, only at inputs in the original domain.
Example: Cancelling a Factor from a Quadratic Quotient
Worked Example: Simplify, Differentiate, and Check
Let \(f(x)=\frac{x^2-5x+6}{x^2-4}\). The original denominator factors as \((x-2)(x+2)\), so the original domain is \(x\ne 2\) and \(x\ne -2\).
Factor the numerator and cancel the common factor, retaining both exclusions:
Differentiate the simpler quotient using the quotient rule. The numerator has derivative \(1\), and the denominator has derivative \(1\):
To check, apply the quotient rule to the original polynomials. Set \(P(x)=x^2-5x+6\) and \(Q(x)=x^2-4\), so \(P'(x)=2x-5\) and \(Q'(x)=2x\). Then
The result matches the derivative of the simplified expression. The input \(x=2\) remains excluded even though the final derivative formula has a value there. The input \(x=-2\) is also excluded, and the final derivative formula has no value there because its denominator is zero.
A Second Pattern: A Repeated Denominator Factor
Cancellation is useful even when the denominator contains a repeated factor. It can reduce the power of the denominator and make the quotient rule calculation shorter. The original exclusions still need to be retained.
Worked Example: Cancelling from a Squared Denominator
Consider \(g(x)=\frac{x^2-1}{(x-1)^2}\). The original denominator is zero at \(x=1\), so the domain is \(x\ne 1\). Factor the numerator and cancel one factor \(x-1\):
Differentiate the simplified quotient. The numerator derivative is \(1\), and the denominator derivative is \(1\):
For a check, use the original numerator \(u=x^2-1\) and denominator \(v=(x-1)^2\). Then \(u'=2x\) and \(v'=2(x-1)\). The quotient rule gives
The quotient-rule result agrees with the simplified calculation. The derivative is not defined at \(x=1\), because \(g\) itself is not defined there.
When Simplifying Helps—and When It Does Not
Simplifying first is especially helpful when a common factor makes the expression much shorter. It may reduce the amount of expansion, the number of terms in the quotient-rule numerator, or the powers appearing in the denominator. If there is no common factor, there may be no useful cancellation, and applying the quotient rule as written is appropriate.
Cancellation only applies to common factors, not to terms that happen to appear in both the numerator and denominator. For example, a quotient of sums cannot be simplified by crossing out matching terms:
The numerator and denominator are sums, not products with a common factor. A safe procedure is to factor the numerator and denominator completely, identify an actual common factor, and then cancel that factor while keeping the original domain restrictions.
A Reliable Procedure
For a rational expression that may simplify, use the following sequence. Keeping the original domain visible prevents a correct algebraic simplification from leading to an incorrect claim about where the original function exists.
Set the original denominator unequal to zero and record every excluded input.
Look for common factors, rather than matching terms within sums.
State that the cancellation is valid on the original domain; do not discard the original exclusions.
Use the applicable differentiation rules and keep the original domain in the conclusion.
Apply the quotient rule to the original expression or substitute a permitted input into both derivative formulas.
Why the Domain Distinction Matters
A simplified expression can describe an extension of the original function: it assigns values at some inputs where the original quotient was undefined. This distinction is useful because it separates two questions. One asks for a shorter formula agreeing with the original function on its domain. The other asks whether the original function itself has a value, or a derivative, at an excluded input. Cancellation answers the first question; by itself, it does not answer the second.
In particular, a derivative formula obtained after cancellation must be read together with its domain. A formula may be algebraically meaningful at an excluded point, but the derivative of the original function is not defined there unless the original function is defined there and the derivative limit exists. In the examples above, the derivative statements were therefore restricted to the original domains.
Check Your Understanding
For each question, keep track of the original domain before simplifying.
- For \(\frac{x^2-16}{x-4}\), what is the original domain, what is the simplified expression on that domain, and what is its derivative?
- Why does cancelling a common factor not make an originally excluded input part of the original function’s domain?
- For \(\frac{x^2-1}{x^2+x}\), identify the original excluded inputs and simplify by factoring.
- What is wrong with cancelling the \(x\) terms in \(\frac{x+3}{x+7}\)?
- State why the derivative of a simplified formula agrees with the derivative of the original quotient at inputs in the original domain.