Why Check the Domain First?
A formula can look smooth and still fail to define a function at particular inputs. Before deciding where a polynomial or rational function is differentiable, first ask where the function is defined.
Every polynomial is defined for every real input. A rational formula, which is a quotient of polynomials, is defined only where its denominator is nonzero. At any real number outside the domain, the ordinary derivative cannot exist: the difference quotient requires a function value at that number.
The important distinction is between a formula that is undefined at a point and a function that has a value there. Algebraic cancellation can simplify values at nearby points, but it does not automatically assign a value at a point excluded by the original formula.
Polynomials: Differentiable Everywhere
A polynomial has the form \(p(x)=c_0+c_1x+\cdots+c_nx^n\), where the coefficients are real numbers and \(n\) is a nonnegative integer. It is defined at every real number. In fact, it is differentiable at every real number as well.
The key reason is that the difference quotient of each power has a finite limit. For a positive integer \(k\), factor the difference of powers:
As \(h\) approaches zero, each of the \(k\) terms approaches \(a^{k-1}\). The quotient therefore approaches \(k a^{k-1}\). For a constant term, the difference quotient is zero. A polynomial is a finite sum of these terms, so the limit of its difference quotient is finite at every real \(a\).
This conclusion is about where a polynomial is differentiable, not a requirement to find its derivative formula. The proof shows why no real input creates a domain problem or an infinite difference quotient.
Worked Example: Checking a Polynomial at a Particular Input
Let \(p(x)=2x^4-3x+1\). Determine where it is differentiable, then check the difference quotient at \(a=-1\).
Because \(p\) is a polynomial, it is defined and differentiable for every real \(x\). To check the specified point, first find \(p(-1)\):
Expand the value at \(-1+h\):
For \(h\ne0\), form the difference quotient and simplify:
Taking the limit as \(h\to0\) gives \(-11\). As a check, the constant term \(6\) cancels in the quotient, and every remaining term except \(-11\) contains a factor of \(h\), so those terms approach zero. Thus \(p\) is differentiable at \(-1\), consistent with the conclusion that it is differentiable for all real \(x\).
Rational Functions: Check the Denominator
A rational formula has the form \(f(x)=p(x)/q(x)\), where \(p\) and \(q\) are polynomials. Its domain consists of the real numbers \(x\) for which \(q(x)\ne0\). The zeros of the denominator must be found before making a claim about differentiability.
At every point in that domain, the rational function is differentiable. A short difference-quotient argument explains the result. Suppose \(q(a)\ne0\). Polynomials \(p\) and \(q\) are differentiable at \(a\), and \(q(x)\) approaches the nonzero value \(q(a)\) as \(x\) approaches \(a\). For sufficiently small \(h\), \(q(a+h)\) is also nonzero. Then:
As \(h\to0\), the two difference quotients in the numerator approach the finite derivatives \(p'(a)\) and \(q'(a)\), and the denominator approaches \(q(a)^2\), which is nonzero. Thus the rational difference quotient has a finite limit. This proves differentiability wherever the original denominator is nonzero.
The theorem identifies the differentiability set directly from the formula's domain. It does not claim that a rational function is differentiable at a denominator zero, even if algebraic simplification produces a simpler expression nearby.
Worked Example: Finding the Differentiability Set from the Denominator
Consider $$ f(x)=\frac{x+1}{x^2-4}. $$ Find the domain and the set of points where \(f\) is differentiable.
Factor the denominator and find its zeros:
The formula is undefined at both zeros, so its domain is \(\mathbb{R}\setminus\{-2,2\}\). At every other real number, the denominator is nonzero. The rational-function result therefore gives the differentiability set:
As a check, \(x=0\) is in this set because \(0^2-4=-4\ne0\); indeed \(f(0)=1/(-4)=-\tfrac14\) is defined. At \(x=2\), the denominator is \(4-4=0\), while the numerator is \(3\), so the original formula has no value there. The same happens at \(x=-2\): the denominator is zero and the numerator is \(-1\). Neither excluded point can be a differentiability point of this function as given.
Cancellation Does Not Restore an Excluded Input
When a numerator and denominator share a factor, cancellation gives an expression with the same values only at inputs where the original quotient was defined. A removed factor can reveal that a hole is removable, but it does not fill that hole in the original function.
Worked Example: A Cancelled Factor and a Removable Hole
Consider $$ f(x)=\frac{x^2-1}{x^2-x}. $$ Find the original domain and differentiability set. Then examine what happens if the hole is filled.
Factor the numerator and denominator:
The denominator is zero when \(x=0\) or \(x=1\), so the original domain is \(\mathbb{R}\setminus\{0,1\}\). On that domain, cancellation is valid:
The simplified expression confirms that \(f\) is differentiable at every point of its original domain, and therefore its differentiability set is \(\mathbb{R}\setminus\{0,1\}\). In particular, cancellation does not make \(f(1)\) defined.
Now define a new function \(F\) by filling the hole: let \(F(x)=f(x)\) for \(x\ne1\), and set \(F(1)=2\). Near \(1\), its values follow \((x+1)/x\). For \(h\ne0\) close to zero:
As \(h\to0\), this approaches \(-1\), so the extended function \(F\) is differentiable at \(1\). The check also confirms the distinction: \(F(1)\) is defined, but the original \(f(1)\) is not. The extension is a different function with an enlarged domain.
A Reliable Domain-and-Differentiability Check
For a polynomial or rational formula, use the domain to organize the decision. The steps below keep the original function separate from any expression obtained by cancellation.
A polynomial has no denominator to check. A rational formula has a polynomial denominator.
For a rational formula, solve \(q(x)=0\). Exclude every real solution from the domain of the original formula.
A polynomial is differentiable on all of \(\mathbb{R}\). A rational function is differentiable at every point in its domain.
Cancellation may describe the values at nearby domain points, but it does not assign a value at an excluded input. Assess a filled-in function separately.
Common Mistake: Treating a Simplified Formula as the Same Domain
A frequent error is to cancel a factor and then include the input where that factor was zero. For example, someone might write:
The conclusion is wrong for the original formula. At \(x=1\), its denominator is zero, so the original function has no value. The derivative definition requires that value, and the original function is not differentiable there.
The correction is to keep the original restriction: the quotient equals \(x+1\) only for \(x\ne1\). If a new function is defined by assigning the value \(2\) at \(x=1\), that extended function is differentiable there, but it is not the same function as the original quotient on the enlarged domain.
Why This Check Matters
Determining where a formula is differentiable begins with a precise domain statement. This prevents two opposite errors: overlooking points where a rational formula is undefined, and incorrectly treating an excluded point as part of the function after cancellation.
When a question asks for a derivative at a specified input, first verify that the input belongs to the function's domain. If it does, polynomial or rational form guarantees differentiability there. If it does not, the original function has no derivative there; a separately defined extension would need to be considered on its own.
Check Your Understanding
For each question, pay attention to the original formula and its domain.
- On what set is \(p(x)=5x^3-x+7\) differentiable?
- Find the domain and differentiability set of \(g(x)=\dfrac{x^2+3}{x^2-9}\).
- For \(h(x)=\dfrac{x^2-4}{x-2}\), does cancellation make the original function differentiable at \(x=2\)? Explain.
- Suppose a rational formula has a denominator that is nonzero at \(a\). What guarantees that it is differentiable at \(a\)?
- How does the differentiability set change if a new function is defined by filling a removable hole? Why is that a different function?
