Let `g(x)=xf(x)`, where `f(x)={{:(x^(2)sin.(1)/(x),":",x ne0),(0,":",x=0):}`. At `x=0`,

Options

A. g is differentiable but g' is not
B. g is differentiable while f is not differentiable
C. both f and g are non differentiable
D. g is differentiable and g' is continuous

Answer

To solve the problem, we need to analyze the function \( g(x) = x f(x) \) where \( f(x) \) is defined as:

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

### Step 1: Define \( g(x) \)

We can express \( g(x) \) for both cases:

1. For \( x \neq 0 \):
\[ g(x) = x f(x) = x \cdot x^2 \sin\left(\frac{1}{x}\right) = x^3 \sin\left(\frac{1}{x}\right) \]

2. For \( x = 0 \):
\[ g(0) = 0 \cdot f(0) = 0 \]

Thus, we have:
\[ g(x) = \begin{cases} x^3 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

### Step 2: Find the derivative \( g'(x) \)

To find \( g'(x) \) for \( x \neq 0 \), we will use the product rule:

\[ g'(x) = \frac{d}{dx}(x^3) \cdot \sin\left(\frac{1}{x}\right) + x^3 \cdot \frac{d}{dx}\left(\sin\left(\frac{1}{x}\right)\right) \]

Calculating each part:

1. The derivative of \( x^3 \) is \( 3x^2 \).
2. For \( \sin\left(\frac{1}{x}\right) \), using the chain rule:
\[ \frac{d}{dx}\left(\sin\left(\frac{1}{x}\right)\right) = \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = -\frac{\cos\left(\frac{1}{x}\right)}{x^2} \]

Now substituting these into the product rule gives:

\[ g'(x) = 3x^2 \sin\left(\frac{1}{x}\right) + x^3 \left(-\frac{\cos\left(\frac{1}{x}\right)}{x^2}\right) \]

This simplifies to:

\[ g'(x) = 3x^2 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right) \]

### Step 3: Find \( g'(0) \)

To find \( g'(0) \), we need to evaluate the limit:

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

Substituting \( g(h) \):

\[ g'(0) = \lim_{h \to 0} \frac{h^3 \sin\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h^2 \sin\left(\frac{1}{h}\right) \]

Since \( |\sin\left(\frac{1}{h}\right)| \leq 1 \), we have:

\[ |h^2 \sin\left(\frac{1}{h}\right)| \leq |h^2| \]

As \( h \to 0 \), \( h^2 \to 0 \). Therefore:

\[ g'(0) = 0 \]

### Step 4: Continuity and Differentiability

1. \( g'(x) \) is defined for \( x \neq 0 \) as \( 3x^2 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right) \).
2. At \( x = 0 \), we found \( g'(0) = 0 \).

To check if \( g'(x) \) is continuous at \( x = 0 \):

\[ \lim_{x \to 0} g'(x) = \lim_{x \to 0} \left(3x^2 \sin\left(\frac{1}{x}\right) - x \cos\left(\frac{1}{x}\right)\right) \]

Both terms approach \( 0 \) as \( x \to 0 \), hence:

\[ \lim_{x \to 0} g'(x) = 0 = g'(0) \]

Thus, \( g'(x) \) is continuous at \( x = 0 \).

### Conclusion

- \( g(x) \) is differentiable at \( x = 0 \) and \( g'(0) = 0 \).
- The function \( g(x) \) is differentiable everywhere, but \( f(x) \) is not differentiable at \( x = 0 \).