Why a Canceled Denominator Zero Needs Care
A rational expression can simplify to a familiar formula even when the original function is undefined at one input. That input is a removable hole: the simplified expression describes the function everywhere in its original domain, but cancellation does not put the missing input back into that domain.
This distinction matters when finding derivatives. At inputs where the original function is defined, its derivative can be found by differentiating the simplified expression. At a hole, however, the original function has no derivative: the derivative requires the function to be defined at that input. A related function obtained by filling in the hole may have a derivative there.
The Domain and the Simplified Formula
Consider a function of the form \(f(x)=\frac{P(x)}{x-a}\), where \(P\) is a polynomial and \(a\) is a real number. Its original domain excludes \(a\), because \(x-a=0\) when \(x=a\). If \(P(a)=0\), then the factor theorem gives a polynomial \(Q\) such that \(P(x)=(x-a)Q(x)\). For every \(x\ne a\), cancellation gives \(f(x)=Q(x)\).
On the original domain, the function agrees with \(Q\). Thus, wherever \(x\ne a\), its derivative is \(Q'(x)\). This uses the standard derivative formulas for polynomials. It does not imply that \(f'(a)\) exists: \(f(a)\) is not defined, so the derivative of \(f\) at \(a\) is not defined either.
We can instead define an extended function \(F\) by using the same values as \(f\) away from \(a\), and setting \(F(a)=Q(a)\). This extension has no hole at \(a\). If \(Q\) is differentiable, then the derivative of the extension at \(a\) is \(Q'(a)\).
Why the Extension Has the Derivative of the Simplified Formula
The claim about the extension follows directly from the definition of derivative. Assume \(P(x)=(x-a)Q(x)\), and define \(F(x)=Q(x)\) for all real \(x\). For any nonzero \(h\), the input \(a+h\) is not \(a\), so the original rational function is defined there and \(f(a+h)=Q(a+h)\). Also, \(F(a)=Q(a)\).
As \(h\) approaches zero, the expression on the right approaches \(Q'(a)\), by the definition of the derivative of \(Q\). Therefore \(F'(a)=Q'(a)\). This proves the derivative formula for the extension. It does not prove that \(f'(a)\) exists: the difference quotient for \(f\) at \(a\) cannot even be formed because \(f(a)\) is missing.
Worked Examples
Worked Example: A Hole at \(x=1\)
Find the derivative of \(f(x)=\frac{x^2-1}{x-1}\) on its domain, and determine whether the original function has a derivative at \(x=1\). Then find the derivative at \(1\) of the extension that fills the hole.
The denominator is zero at \(x=1\), so the original domain is \(x\ne 1\). Factor the numerator:
For \(x\ne 1\), cancellation gives \(f(x)=x+1\). Differentiating this simplified expression gives \(f'(x)=1\) for \(x\ne 1\). This derivative formula applies on the original domain only.
Define the extension \(F(x)=x+1\) for every real \(x\). Then \(F(1)=1+1=2\). Using the difference quotient,
Thus \(F'(1)=1\). As a check, differentiating the extended formula \(F(x)=x+1\) gives \(F'(x)=1\), which agrees at \(x=1\). The original function \(f\), however, has no value at \(1\), so \(f'(1)\) is undefined.
Worked Example: A Different Hole at \(x=1\)
Consider \(g(x)=\frac{x^2-4x+3}{x-1}\). Find the derivative on the original domain and the derivative at the hole of the extension that fills it.
The denominator vanishes at \(x=1\), so \(g\) is defined only when \(x\ne 1\). Factor the numerator by finding two numbers whose product is \(3\) and whose sum is \(-4\), namely \(-1\) and \(-3\):
For \(x\ne 1\), cancel the common factor to obtain \(g(x)=x-3\). Therefore \(g'(x)=1\) for every \(x\ne 1\). The original function is still undefined at \(1\).
Define the extension \(G(x)=x-3\) for all real \(x\). Its value at the hole is \(G(1)=1-3=-2\). The difference quotient at that input is
Hence \(G'(1)=1\). As a check, differentiating \(G(x)=x-3\) gives \(G'(x)=1\), including at \(x=1\). This verifies the extension’s derivative, not a derivative of \(g\) at the excluded input.
Worked Example: A Cubic Numerator
Let \(h(x)=\frac{x^3-8}{x-2}\). Find its derivative on its original domain and the derivative at \(x=2\) of the extension that fills the hole.
The original function is undefined when \(x=2\). Factor the difference of cubes:
For \(x\ne 2\), this gives \(h(x)=x^2+2x+4\). Differentiating the polynomial yields
Define the extension \(H(x)=x^2+2x+4\) for all real \(x\). Its value at the hole is
To check the derivative of the extension at \(2\) from its definition, substitute \(2+h\) into the polynomial:
Therefore, for \(h\ne 0\),
As \(h\) approaches zero, \(6+h\) approaches \(6\), so \(H'(2)=6\). This agrees with differentiating \(H\): \(H'(x)=2x+2\), and \(H'(2)=2(2)+2=6\). The original function \(h\) is undefined at \(2\), so it has no derivative there.
A Common Mistake: Letting Cancellation Change the Domain
A tempting but incorrect calculation is to cancel a factor and then act as though the canceled input belongs to the original function’s domain. For example, after simplifying \(\frac{x^2-1}{x-1}\) to \(x+1\), one might write:
The first equality is not valid for the original function: its denominator is zero at \(1\), so \(f(1)\) is undefined. The value \(2\) belongs to the extension \(F(x)=x+1\), not to \(f\). The corrected conclusions are \(f'(x)=1\) for \(x\ne 1\), \(f'(1)\) is undefined, and \(F'(1)=1\) for the extension.
How to Handle a Canceled Denominator Zero
When a denominator zero may be removable, keep the algebra and the domain separate. The process is:
Set the original denominator equal to zero. Record that the original function is not defined at any such input.
If the numerator and denominator share a factor, simplify the expression while retaining the original exclusion.
The resulting formula gives the derivative of the original function at inputs where the original function is defined.
If the simplified expression is defined at the excluded input, it defines an extension there. Its derivative at that input belongs to the extension, not the original function.
The distinction is useful whenever algebra makes a function look simpler than its original definition. A simplified formula can describe all the function’s existing values while also suggesting a natural value for a missing point. Those are two different statements: the first concerns equality on the original domain; the second defines a new function by filling the hole.
Check Your Understanding
For each question, distinguish carefully between the original function and any extension that fills its hole.
- For \(r(x)=\frac{x^2-4}{x-2}\), what is the original domain, and what formula equals \(r(x)\) on that domain?
- What is the derivative of \(r\) at inputs in its original domain?
- If \(R(x)=x+2\) for every real \(x\), including \(x=2\), what is \(R'(2)\)?
- Why does the answer to the previous question not give \(r'(2)\)?
- For \(s(x)=\frac{x^2-9}{x-3}\), state the derivative of \(s\) for \(x\ne 3\), and explain whether \(s'(3)\) is defined.
