Problem 39 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

2-39. Use the function \(f: \mathbf{R} \rightarrow \mathbf{R}\) defined by $$ f(x)= \begin{cases}\frac{x}{2}+x^{2} \sin \frac{1}{x} & x \neq 0 \\ 0 & x=0\end{cases} $$ to show that continuity of the derivative cannot be eliminated from the hypothesis of Theorem \(2-11 .\)

Step-by-Step Solution

Answer

Function has a discontinuous derivative at x=0, contradicting Theorem 2-11.

Step 1 - Define the function

Rewrite the given function for clarity: \[ f(x) = \begin{cases} \frac{x}{2} + x^{2} \sin \frac{1}{x} & x eq 0 \ 0 & x = 0 \end{cases} \]

Step 2 - Verify differentiability at non-zero

For \(x eq 0\), compute the derivative using the product rule and chain rule: \[ f'(x) = \frac{1}{2} + \frac{d}{dx} \left( x^{2} \sin \frac{1}{x} \right) \] Use the product rule \( (uv)' = u'v + uv' \): \[ f'(x) = \frac{1}{2} + 2x \sin \frac{1}{x} + x^2 \left( \cos \frac{1}{x} \cdot -\frac{1}{x^2} \right) \] Simplify: \[ f'(x) = \frac{1}{2} + 2x \sin \frac{1}{x} - \cos \frac{1}{x} \]

Step 3 - Evaluate derivative at zero

Evaluate the limit definition of the derivative at \(x = 0\): \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \] Since \( f(0) = 0 \): \[ f'(0) = \lim_{h \to 0} \frac{\frac{h}{2} + h^2 \sin \frac{1}{h}}{h} \] Simplify the fraction: \[ f'(0) = \lim_{h \to 0} \left( \frac{1}{2} + h \sin \frac{1}{h} \right) \] Since \( h \sin \frac{1}{h} \) oscillates between -1 and 1, so the limit is \( \frac{1}{2} \). Thus, \( f'(0) = \frac{1}{2} \).

Step 4 - Check continuity of the derivative

For continuity, check the limit of the derivative as \( x \to 0 \): \[ \lim_{x \to 0} f'(x) = \lim_{x \to 0} \left( \frac{1}{2} + 2x \sin \frac{1}{x} - \cos \frac{1}{x} \right) \] \( 2x \sin \frac{1}{x} \to 0 \) but \( \cos \frac{1}{x} \) oscillates between -1 and 1. Therefore, \( f'(x) \) does not approach \( \frac{1}{2} \) continuously. Hence, \( f'(x) \) is not continuous at \( x = 0 \).

Step 5: Conclusion - Relate to Theorem 2-11

Theorem 2-11 likely involves conditions sufficient for certain properties when the derivative is continuous. Since \( f'(x) \) is not continuous at \(x = 0\), this function serves as a counterexample to show that continuity of the derivative is crucial.

Key Concepts

Differentiable Functions

Differentiable functions are those that have a well-defined derivative at every point in their domain. This means that the function has a tangent that is not only existing but also smooth without any sharp breaks. For a function to be differentiable at a point, it must also be continuous at that point.
For example, if we consider the function from the original problem: \[ f(x) = \begin{cases} \frac{x}{2} + x^{2} \sin \frac{1}{x} & x eq 0 \ 0 & x = 0 \end{cases} \] Its differentiability at non-zero values relies on the product and chain rules, while at zero, it requires us to use the limit definition of the derivative. This ensures that the slopes of the secant lines approach a single value at the point of interest.

Product Rule

The product rule is crucial for finding the derivatives of the products of two functions. If you have functions \(u(x)\) and \(v(x)\), the product rule states that: \[ (uv)' = u'v + uv' \]In the given exercise, we use this rule to differentiate the term \(x^{2} \sin \frac{1}{x}\). Here's a detailed application:

• First, identify \(u = x^2\) and \(v = \sin \frac{1}{x}\).
• Next, calculate their derivatives: \(u' = 2x\) and \(v' = \cos \frac{1}{x} \times -\frac{1}{x^2}\).
• Finally, apply the product rule: \[ \frac{d}{dx} \left( x^2 \sin \frac{1}{x} \right) = 2x \sin \frac{1}{x} + x^{2} \left( \cos \frac{1}{x} \times -\frac{1}{x^2}\right) \]

This simplifies the derivative of the function for non-zero values of \(x\).

Chain Rule

The chain rule is used for finding the derivative of composite functions. If you have a function composed as \(f(g(x))\), the chain rule states: \[ (f(g(x)))' = f'(g(x)) \, g'(x) \]In the exercise, let's use the chain rule for the term \( x^2 \sin \frac{1}{x} \):

• Consider \(g(x) = \frac{1}{x}\).
• Then, \(\sin \frac{1}{x} = \sin (g(x))\).
• The inner function's derivative is \( g'(x) = -\frac{1}{x^2} \).

Combining these results with the product rule gives the complete derivative for functions involving more complex inner compositions.

Limit Definition of the Derivative

The limit definition of the derivative expresses how a function approaches a certain value as we move infinitesimally close to a point. This is essential for finding the derivative at points where direct computation might be problematic, like at \(x = 0\) in our exercise. The definition is: \[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]For the function \( f(x) \) at \(x = 0\), this becomes: \[ f'(0) = \lim_{h \to 0} \frac{\frac{h}{2} + h^2 \sin \frac{1}{h}}{h} = \lim_{h \to 0} \left( \frac{1}{2} + h \sin \frac{1}{h} \right) \]Since \(h \sin \frac{1}{h}\) oscillates between -1 and 1, the limit simplifies to \(\frac{1}{2}\), showing that the derivative at zero exists and equals \(\frac{1}{2}\).