Debunking the Myth: Continuous Functions and Their Derivatives

Many people believe that if a function is continuous, then it must also be differentiable, and its derivative must also be continuous. This belief, however, is not entirely true, as we shall see through the exploration of specific examples and proofs.

Introduction

At first glance, it may seem logical to assume that the derivative of a continuous function should also be continuous. However, this assumption is not always valid, as illustrated by several counterexamples. In this article, we will explore the reasons why this statement is not universally true and provide a detailed proof using a specific function.

Counterexample: A Function Without a Continuous Derivative

Consider the function f(x) x^2 sin(1/x) for x ≠ 0 and f(0) 0. This function is continuous everywhere, but it is not differentiable at x 0. Let's explore this in more detail.

Derivatives of the Function

For x ≠ 0, the function f(x) has the derivative:

f'(x) 2x sin(1/x) - cos(1/x)

At x 0, we can find the derivative using the limit definition:

f'(0) lim_{h → 0} frac{f(h) - f(0)}{h} lim_{h → 0} frac{h^2 sin(1/h)}{h} lim_{h → 0} h sin(1/h) 0

Thus, the derivative of f(x) at x 0 exists and is equal to 0. However, the derivative f'(x) is not continuous at x 0. To verify this, let's compute lim_{x → 0} f'(x) from both sides:

lim_{x → 0^} f'(x) lim_{x → 0^} (2x sin(1/x) - cos(1/x))

lim_{x → 0^} 2x sin(1/x) - lim_{x → 0^} cos(1/x)

The term 2x sin(1/x) approaches 0 as x approaches 0, while cos(1/x) does not approach a limit due to its oscillatory behavior. Similarly, we can show:

lim_{x → 0^-} f'(x) lim_{x → 0^-} (2x sin(1/x) - cos(1/x))

lim_{x → 0^-} 2x sin(1/x) - lim_{x → 0^-} cos(1/x)

Since both limits of 2x sin(1/x) approach 0 from both sides, the limit of cos(1/x) does not exist due to its oscillation. Therefore, lim_{x → 0} f'(x) does not exist, and f'(x) is not continuous at x 0.

Another Counterexample

Let's consider another function: g(x) x. This function is continuous at x 0, but its derivative does not exist at this point. We will prove this step-by-step.

Proof of Continuity of the Function g(x) at x 0

To show that g(x) x is continuous at x 0, we need to show:

lim_{x → 0} g(x) g(0)

which means:

lim_{x → 0} x 0

This limit is straightforward and can be proved as follows: For every epsilon > 0, we can choose delta epsilon. Then for all x such that |x - 0| , it holds that:

|x|

Therefore, the function g(x) x is continuous at x 0.

Proof that the Derivative of g(x) at x 0 Does Not Exist

The derivative of g(x) at x 0 is defined as:

lim_{h → 0} frac{g(0 h) - g(0)}{h} lim_{h → 0} frac{h - 0}{h} lim_{h → 0} 1 1

From the right side (h → 0^):

lim_{h → 0^} frac{h}{h} lim_{h → 0^} 1 1

However, from the left side (h → 0^):

lim_{h → 0^-} frac{h}{h} lim_{h → 0^-} -1 -1

Since the limit from the right side is not equal to the limit from the left side, the derivative of g(x) at x 0 does not exist.

Conclusion

In conclusion, it is evident that not every continuous function has a continuous derivative. Both examples presented in this article, f(x) x^2 sin(1/x) and g(x) x, demonstrate that the derivative of a continuous function can be discontinuous at certain points. A continuous function might have a derivative at every point, but the derivative may not be continuous. Therefore, the original statement is incorrect.

References and Further Reading

This article has covered important concepts related to continuous functions and their derivatives. For further reading on this topic, consider exploring:
Continuous Functions and Their Properties The Intermediate Value Property Differentiability and Continuity