What Does It Mean to Check a Claimed Derivative?
A derivative value is not established by a plausible-looking formula or by testing a few nearby inputs. At a point, the derivative is defined by a limit. To check a claim, form that limit, simplify the quotient where possible, and determine whether its limit is the claimed number.
Let \(f\) be defined on an open interval containing \(a\). For nonzero \(h\) sufficiently close to zero, the difference quotient at \(a\) is
The condition \(h\ne 0\) matters: the quotient is not evaluated at \(h=0\). Instead, we simplify it for nonzero \(h\), then take the limit as \(h\) approaches zero. The expressions after simplification may have a value or a limit at zero even though the original quotient is undefined there.
This tutorial considers points inside the domain, where the limit is two-sided. At an endpoint of a domain, the relevant derivative definition uses the appropriate one-sided limit.
A Reliable Test
Suppose someone claims that \(f'(a)=c\). The claim asks whether the difference quotient approaches \(c\) as \(h\to0\). It is often useful to simplify the quotient first and then compare the resulting limit with \(c\). If the quotient approaches a different number, the claim is false; if it has no finite limit, there is no derivative value of the claimed kind.
Substitute \(a+h\) and \(a\) into the function, in that order, and divide their difference by \(h\).
Use valid algebra, such as factoring a common factor or multiplying by a conjugate. Keep track of signs.
Evaluate the limit of the simplified expression as \(h\to0\).
The proposed value is confirmed only if it equals the limit just computed.
There is a simple reason this comparison is decisive: a function cannot have two different limits at the same point. Thus, once the difference quotient is shown to approach a particular number, any different proposed derivative value is ruled out.
Worked Examples
Worked Example: Confirming a Derivative of a Square-Root Function
Let \(f(x)=\sqrt{x}\), and check the claim \(f'(9)=\frac{1}{6}\).
Start with the definition. Since \(f(9)=3\),
For \(h\ne0\), multiply numerator and denominator by the conjugate \(\sqrt{9+h}+3\):
The cancellation of \(h\) is valid because the difference quotient is considered only for \(h\ne0\). Now take the limit:
The limit equals the proposed value, so the claim is confirmed: \(f'(9)=\frac{1}{6}\). Substituting \(h=0\) into the original quotient would have produced \(0/0\); the valid check instead simplifies for nonzero \(h\) and then takes the limit.
Worked Example: Checking a Reciprocal Function and Its Sign
Let \(f(x)=\frac{1}{x}\), and check the claim \(f'(-2)=-\frac{1}{4}\). The function is defined near \(-2\), so the difference quotient is appropriate. Since \(f(-2)=-\frac{1}{2}\),
In the last line, \(h\) cancels because \(h\ne0\). Taking the limit gives
This confirms the claim. It also rejects the alternative claim \(f'(-2)=\frac{1}{4}\): the difference quotient approaches \(-\frac{1}{4}\), not \(\frac{1}{4}\). The negative sign comes from the denominator \(h-2\), whose limit is \(-2\).
Worked Example: Rejecting a Sign Error for a Polynomial
Let \(f(x)=x^2\). A proposed value is \(f'(-2)=4\). Compute the quotient to test it. First, \(f(-2)=4\), so
Therefore,
The proposed value \(4\) is false: the actual limit is \(-4\). As a check on the calculation, substituting \(h=0\) into the simplified expression \(-4+h\) gives \(-4\), exactly the limit found. This does not mean that \(h=0\) can be substituted into the original quotient; that quotient is defined only for \(h\ne0\).
Why the Limit Settles the Claim
The comparison above rests on uniqueness of limits. Here is a short justification. Suppose the difference quotient \(Q(h)\) has limit \(L\) as \(h\to0\), and also has limit \(c\). If \(L\ne c\), let \(\varepsilon=\frac{|L-c|}{3}\), which is positive. By the definition of the two limits, for all sufficiently small nonzero \(h\), both \(|Q(h)-L|<\varepsilon\) and \(|Q(h)-c|<\varepsilon\) hold. The triangle inequality then gives
which is impossible for a positive number \(|L-c|\). Therefore \(L=c\). In particular, if simplifying the difference quotient shows that its limit is \(L\), a claimed value different from \(L\) cannot also be the derivative.
This is why the final comparison is not merely a numerical check. It follows from the definition of derivative and the uniqueness of a limit. When the limit exists, its value is the derivative; there is no second value that could also satisfy the definition.
A Common Pitfall: Canceling Without Preserving the Sign
Cancellation can be essential, but it must be applied to a common factor, and the other factors must be carried along correctly. Consider the reciprocal example. A sign error might turn its numerator into \(2-(h-2)\) instead of the correct \(2+(h-2)\), producing
This is not an equivalent expression for the original quotient. The original numerator is \(\frac{1}{h-2}+\frac{1}{2}\), so using the common denominator \(2(h-2)\) gives \(2+(h-2)=h\), not \(2-(h-2)\). The correct simplification is
Its limit is \(-\frac{1}{4}\), as required. The incorrect numerator would not even contain a factor of \(h\), so canceling \(h\) from it would be invalid. Before canceling, check that the entire numerator and denominator contain the same factor, and retain any negative signs in the remaining expression.
How to Present a Convincing Check
A clear verification or refutation should make three facts visible: the original difference quotient, the algebra used to simplify it for \(h\ne0\), and the limit that is compared with the claimed value. This makes it possible to locate a sign error or an invalid cancellation rather than hiding it inside a final answer.
For a verification, end by stating that the computed limit equals the proposed number. For a refutation, state the actual limit and explicitly note that it differs from the proposed number. If the quotient has no finite limit, explain that the derivative does not exist as a finite value at that point; a proposed finite value is then false.
Check Your Understanding
For each question, use the difference quotient and show enough algebra to justify the limit.
- For \(f(x)=\sqrt{x+7}\), check the claim \(f'(2)=\frac{1}{6}\).
- For \(f(x)=\frac{1}{x}\), test the claim \(f'(1)=-1\). Identify the sign that determines the limit.
- For \(f(x)=x^2\), compute the difference quotient at \(a=3\) and decide whether the proposed value \(6\) is correct.
- Why is it valid to cancel \(h\) after factoring it from the difference quotient, even though the derivative is defined by a limit as \(h\to0\)?
- If a simplified difference quotient has limit \(-2\), can the derivative at that point also equal \(2\)? Explain using uniqueness of limits.
