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=0

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 \(x = 0\)

To check if \(f(x)\) is differentiable at \(x = 0\), we need to compute the right-hand and left-hand derivatives at that point.

**Right-hand derivative:**

The formula for the right-hand derivative at \(x = 0\) is given by:

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

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 have:

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

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

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

Thus, the right-hand derivative at \(x = 0\) is:

\[ f'(0) = 0 \]

**Left-hand derivative:**

The formula for the left-hand derivative at \(x = 0\) is:

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

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

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

By the same reasoning as above, we find:

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

Thus, the left-hand derivative at \(x = 0\) is also:

\[ f'(0) = 0 \]

Since both the right-hand and left-hand derivatives at \(x = 0\) are equal, we conclude that \(f(x)\) is differentiable at \(x = 0\) and \(f'(0) = 0\).

### Step 2: Find the derivative \(f'(x)\) for \(x \neq 0\)

For \(x \neq 0\), we can use the product rule to differentiate \(f(x) = x^2 \sin\left(\frac{1}{x}\right)\):

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

Calculating the derivatives:

1. \(\frac{d}{dx}(x^2) = 2x\)
2. To differentiate \(\sin\left(\frac{1}{x}\right)\), we use the chain rule:

\[ \frac{d}{dx}\left(\sin\left(\frac{1}{x}\right)\right) = \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \]

Thus,

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

This simplifies to:

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

### Step 3: Check continuity of \(f'(x)\) at \(x = 0\)

Now we need to check if \(f'(x)\) is continuous at \(x = 0\). We need to find 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 and does not approach a limit. Therefore, the limit does not exist.

### Conclusion

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