Conisder the function `f(x)={{:(x^(2)ln|x|,xne0),(0,x=0):}. "What is"` f'(0) equal to?

Options

A. 0
B. 1
C. `-1`
D. It does not exist

Explanation

To find \( f'(0) \) for the function defined as:

\[ f(x) = \begin{cases} x^2 \ln |x| & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

we need to calculate the derivative at \( x = 0 \) using the definition of the derivative:

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

### Step 1: Substitute the function into the derivative formula

Since \( f(0) = 0 \), we can substitute this into the derivative formula:

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

### Step 2: Evaluate \( f(h) \) for \( h \neq 0 \)

For \( h \neq 0 \):

\[ f(h) = h^2 \ln |h| \]

Thus, we have:

\[ f'(0) = \lim_{h \to 0} \frac{h^2 \ln |h|}{h} \]

### Step 3: Simplify the expression

This simplifies to:

\[ f'(0) = \lim_{h \to 0} h \ln |h| \]

### Step 4: Evaluate the limit

To evaluate \( \lim_{h \to 0} h \ln |h| \), we can rewrite \( \ln |h| \) as \( \ln h \) (since \( h \) approaches 0 from the positive side):

\[ \lim_{h \to 0^+} h \ln h \]

As \( h \to 0^+ \), \( \ln h \to -\infty \) and \( h \to 0 \). The product \( h \ln h \) approaches 0. We can use L'Hôpital's rule or recognize that:

\[ \lim_{h \to 0^+} h \ln h = 0 \]

### Step 5: Conclude the derivative at \( x = 0 \)

Thus, we find:

\[ f'(0) = 0 \]

### Final Answer

\[ f'(0) = 0 \]

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