What Derivative Information Does Not Guarantee
Taylor’s theorem turns suitable derivative hypotheses into quantitative estimates. The hypotheses matter: differentiability at one point gives a first-order approximation there, but it does not automatically provide a quadratic error bound. Likewise, knowing that a function is bounded does not imply that its derivative is bounded. In this tutorial, we examine examples that make these limits precise.
We will use the definition of differentiability directly. If \(f\) is differentiable at \(a\), then
The condition on \(r(h)\) says that the remainder is small compared with \(|h|\). It does not say that the remainder is bounded by a constant times \(h^2\). That stronger estimate generally requires more information than differentiability at a single point.
A First-Order Approximation Need Not Have a Quadratic Error
The next proposition gives a direct counterexample to the claim that differentiability at a point always produces a quadratic error estimate after the linear term is removed.
Proof. Since \(f(0)=0\), for \(h\ne0\) the difference quotient at zero is
Its absolute value is \(|h|^{1/2}\), which tends to zero as \(h\to0\). Thus \(f\) is differentiable at zero and \(f'(0)=0\). If \(f(x)\) were \(O(x^2)\), there would be constants \(C\geq0\) and \(\delta>0\) such that
Dividing by \(|x|^{3/2}\) would give \(1\leq C|x|^{1/2}\) for every such \(x\). But \(C|x|^{1/2}\to0\) as \(x\to0\), so this inequality cannot hold for all sufficiently small nonzero \(x\). Hence \(f(x)\) is not \(O(x^2)\). \(\square\)
The example has a useful scale interpretation. Its value \(|x|^{3/2}\) is smaller than \(|x|\) near zero, as required for a zero derivative, but larger than every fixed constant times \(|x|^2\) sufficiently close to zero. The first-order approximation is valid; the stronger quadratic estimate is not.
Worked Example: Testing a Proposed Quadratic Estimate
Consider \(f(x)=|x|^{3/2}\) at \(0\). We have \(f(0)=0\) and \(f'(0)=0\), so the first-order Taylor expression is the constant \(0\). Test whether the error can be bounded by \(4x^2\). For \(x\ne0\), the proposed inequality is
This last inequality requires \(|x|\geq1/16\), so it fails for every \(x\) with \(0<|x|<1/16\). In fact, the same argument rules out any fixed constant in place of \(4\). The derivative at the point alone supplies no quadratic bound for this error.
A Bounded Function Can Have an Unbounded Derivative
Derivative bounds and function bounds are different kinds of information. A bound on the derivative can control increments through the Mean Value Theorem. The reverse implication does not hold on an unbounded domain: a function can remain in a fixed range while changing increasingly rapidly.
Proof. Since \(|\sin y|\leq1\) for every real \(y\), we have \(|g(x)|\leq1\) for every \(x\), so \(g\) is bounded. By the chain rule,
For each positive integer \(n\), set \(x_n=(2\pi n)^{1/3}\). Then \(x_n\to\infty\), \(x_n^3=2\pi n\), and \(\cos(x_n^3)=\cos(2\pi n)=1\). Therefore
As \(n\to\infty\), these derivative values tend to \(+\infty\). Thus \(g'\) is unbounded, even though \(g\) is bounded. \(\square\)
The sequence in the proof is essential: to show that a function is unbounded, it is enough to find inputs where its values grow without bound. Here the inputs are chosen so the cosine factor equals \(1\), leaving the growing factor \(3x_n^2\) exposed.
Worked Example: Verifying the Derivative Values
For \(g(x)=\sin(x^3)\), the chain rule gives \(g'(x)=3x^2\cos(x^3)\). Choose \(x_1=(2\pi)^{1/3}\). Substitution gives
For \(x_2=(4\pi)^{1/3}\), the same calculation gives
More generally, \(x_n=(2\pi n)^{1/3}\) gives \(g'(x_n)=3(2\pi n)^{2/3}\). These values increase without bound. At the same time, each function value satisfies \(|g(x_n)|=|\sin(2\pi n)|=0\); indeed, all values of \(g\), not just those on this sequence, lie between \(-1\) and \(1\).
Derivative Values Near a Point Can Behave Very Differently
A derivative at a single point describes difference quotients based at that point. It does not, by itself, bound derivatives at nearby points. The following example places these two facts side by side: the derivative at zero exists and equals zero, while derivative values arbitrarily close to zero have arbitrarily large magnitude.
Worked Example: Zero Derivative at the Origin but Unbounded Nearby Derivatives
Define
At zero, the difference quotient is
Its absolute value is at most \(|h|\), so it tends to zero. Thus \(F'(0)=0\). For \(x\ne0\), the product and chain rules give
Now take \(x_n=(2\pi n)^{-1/2}\), where \(n\) is a positive integer. Then \(x_n\to0\), \(1/x_n^2=2\pi n\), and \(\sin(1/x_n^2)=0\), \(\cos(1/x_n^2)=1\). Substitution into the derivative formula yields
Consequently, \(|F'(x_n)|=2\sqrt{2\pi n}\to\infty\), even though \(F'(0)=0\). The derivative at zero exists, but it does not provide a finite bound for derivative values in any neighborhood of zero.
This example also emphasizes the difference between the quotient that defines \(F'(0)\) and the formula for \(F'(x)\) at \(x\ne0\). The first is controlled by \(|h|\), which tends to zero. The second contains a term proportional to \(1/x\), whose size can grow without bound along a suitable sequence. Neither calculation contradicts the other: they answer different questions.
Using Counterexamples Carefully
A counterexample is most effective when it targets exactly one unsupported implication. In the first example, differentiability and a zero derivative hold, but a quadratic error estimate fails. In the second, bounded function values coexist with an unbounded derivative. In the third, a derivative exists at the origin, but nearby derivative values are not bounded. In each case, direct calculations verify both the claimed property and the failure of the stronger conclusion.
Identify which hypothesis is actually known, such as differentiability at one point or bounded function values.
For a failed error estimate, compare powers of the displacement. For unbounded derivative values, choose inputs that control oscillating factors.
Prove the example has the required regularity, then calculate explicitly why the proposed stronger conclusion fails.
A common pitfall is to confuse a conclusion about one point with a uniform conclusion on a neighborhood. Differentiability at \(a\) is a limit statement about difference quotients based at \(a\); it does not assert a bound on \(f'(x)\) for nearby \(x\). Similarly, boundedness of \(f\) controls its values, not the rate at which those values change.
These constructions motivate a closer study of how derivatives behave as functions in their own right. In particular, a derivative need not vary continuously, even when the original function is differentiable. The examples here show why one should not assume such additional regularity without proving it or imposing it as a hypothesis.
Check Your Understanding
Use the calculations and distinctions in this tutorial to assess each claim.
- Why does differentiability at zero with derivative zero imply an error that is small compared with \(|x|\), but not necessarily an error that is \(O(x^2)\)?
- For \(f(x)=|x|^{3/2}\), what inequality would have to hold if \(f(x)\) were \(O(x^2)\), and why does it fail near zero?
- Which sequence shows that the derivative of \(g(x)=\sin(x^3)\) is unbounded?
- In the example \(F(x)=x^2\sin(1/x^2)\), why does the difference quotient at zero tend to zero while the derivative values along \(x_n=(2\pi n)^{-1/2}\) are unbounded?
- What is the difference between a bound on the values of a function and a bound on its derivative?