Show that the function defined by `f(x) = {{:(x^(2) sin 1//x",",x ne 0),(0",",x = 0):}` is differentiable for every value of x, but the derivative is not continuous for x =

Answer

To show that the function defined by

\[ f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

is differentiable for every value of \( x \), but the derivative is not continuous at \( x = 0 \), we will follow these steps:

### Step 1: Check Differentiability at \( x \neq 0 \)

For \( x \neq 0 \), we can differentiate \( f(x) \) using the product and chain rule.

\[ f'(x) = \frac{d}{dx}\left(x^2 \sin\left(\frac{1}{x}\right)\right) \]

Using the product rule, we have:

\[ f'(x) = 2x \sin\left(\frac{1}{x}\right) + x^2 \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \]

This simplifies to:

\[ f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \]

### Step 2: Check Differentiability at \( x = 0 \)

To check if \( f(x) \) is differentiable at \( x = 0 \), we need to find the limit:

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \]

Substituting \( f(h) \) and \( f(0) \):

\[ f'(0) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right) - 0}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right) \]

Since \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1, we can say:

\[ -h \leq h \sin\left(\frac{1}{h}\right) \leq h \]

As \( h \to 0 \), both bounds approach 0. By the Squeeze Theorem:

\[ \lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0 \]

Thus, \( f'(0) = 0 \).

### Step 3: Show that the Derivative is Not Continuous at \( x = 0 \)

Now we need to check the limit of \( f'(x) \) as \( x \) approaches 0:

\[ \lim_{x \to 0} f'(x) = \lim_{x \to 0} \left(2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\right) \]

The term \( 2x \sin\left(\frac{1}{x}\right) \) approaches 0 as \( x \to 0 \). However, \( \cos\left(\frac{1}{x}\right) \) oscillates between -1 and 1, meaning:

\[ \lim_{x \to 0} f'(x) \text{ does not exist.} \]

Thus, \( f'(x) \) is not continuous at \( x = 0 \).

### Conclusion

The function \( f(x) \) is differentiable for every value of \( x \), but the derivative \( f'(x) \) is not continuous at \( x = 0 \).

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