For the function `f (x) = {{:(x/(1+e^(1//x))", " x ne 0),(" 0 , "x = 0):}`, the derivative from the right, `f'(0^(+)) `= … and the derivative from the left, ` f'(0^(-))` = … .

Explanation

To find the right-hand derivative \( f'(0^+) \) and the left-hand derivative \( f'(0^-) \) of the function

\[ f(x) = \begin{cases} \frac{x}{1 + e^{\frac{1}{x}} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

we will use the definition of the derivative.

### Step 1: Finding the Right-Hand Derivative \( f'(0^+) \)

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

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

Substituting \( f(0) = 0 \):

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

For \( h > 0 \):

\[ f(h) = \frac{h}{1 + e^{\frac{1}{h}}} \]

So we have:

\[ f'(0^+) = \lim_{h \to 0^+} \frac{\frac{h}{1 + e^{\frac{1}{h}}}}{h} = \lim_{h \to 0^+} \frac{1}{1 + e^{\frac{1}{h}}} \]

As \( h \to 0^+ \), \( \frac{1}{h} \to +\infty \), which implies \( e^{\frac{1}{h}} \to +\infty \). Therefore:

\[ f'(0^+) = \lim_{h \to 0^+} \frac{1}{1 + e^{\frac{1}{h}}} = \frac{1}{\infty} = 0 \]

### Step 2: Finding the Left-Hand Derivative \( f'(0^-) \)

The left-hand derivative at \( x = 0 \) is given by:

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

Substituting \( f(0) = 0 \):

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

For \( h < 0 \):

\[ f(-h) = \frac{-h}{1 + e^{\frac{1}{-h}}} \]

Thus:

\[ f'(0^-) = -\lim_{h \to 0^-} \frac{\frac{-h}{1 + e^{\frac{1}{-h}}}}{h} = -\lim_{h \to 0^-} \frac{-1}{1 + e^{\frac{1}{-h}}} \]

As \( h \to 0^- \), \( \frac{1}{-h} \to -\infty \), which implies \( e^{\frac{1}{-h}} \to 0 \). Therefore:

\[ f'(0^-) = -\lim_{h \to 0^-} \frac{-1}{1 + 0} = 1 \]

### Final Results

Thus, we have:

\[ f'(0^+) = 0 \quad \text{and} \quad f'(0^-) = 1 \]