The set of all points where the function `f(x)=sqrt(1-e^(-x^(2)))` is differentiable is :

Options

A. `(0,oo)`
B. `(-oo,oo)`
C. `(-oo,0)uu(0,oo)`
D. `(-1,oo)`

Answer

To determine the set of all points where the function \( f(x) = \sqrt{1 - e^{-x^2}} \) is differentiable, we need to analyze the function and its derivative.

### Step-by-Step Solution:

1. **Identify the function**:
The function is given as:
\[ f(x) = \sqrt{1 - e^{-x^2}} \]

2. **Determine the domain of the function**:
The expression inside the square root, \( 1 - e^{-x^2} \), must be non-negative for \( f(x) \) to be defined.
Since \( e^{-x^2} \) is always positive for all real \( x \), \( 1 - e^{-x^2} \) is non-negative for all \( x \). Therefore, the function is defined for all \( x \in \mathbb{R} \).

3. **Find the derivative**:
To find where the function is differentiable, we need to compute the derivative \( f'(x) \). We can use the chain rule:
\[ f'(x) = \frac{d}{dx} \left(1 - e^{-x^2}\right)^{1/2} \]
Using the chain rule, we have:
\[ f'(x) = \frac{1}{2} (1 - e^{-x^2})^{-1/2} \cdot \frac{d}{dx}(1 - e^{-x^2}) \]
Now, we differentiate \( 1 - e^{-x^2} \):
\[ \frac{d}{dx}(1 - e^{-x^2}) = 0 + e^{-x^2} \cdot (2x) = 2x e^{-x^2} \]
Thus, substituting back, we get:
\[ f'(x) = \frac{1}{2} (1 - e^{-x^2})^{-1/2} \cdot (2x e^{-x^2}) = x e^{-x^2} (1 - e^{-x^2})^{-1/2} \]

4. **Determine where the derivative is undefined**:
The derivative \( f'(x) \) will be undefined if the denominator \( (1 - e^{-x^2})^{-1/2} \) is undefined, which occurs when \( 1 - e^{-x^2} = 0 \). This happens when:
\[ e^{-x^2} = 1 \implies -x^2 = 0 \implies x = 0 \]

5. **Conclusion**:
Therefore, the function \( f(x) \) is differentiable everywhere except at \( x = 0 \). The set of all points where the function is differentiable is:
\[ x \in \mathbb{R} \setminus \{0\} \quad \text{or} \quad (-\infty, 0) \cup (0, \infty) \]

### Final Answer:
The set of all points where the function \( f(x) \) is differentiable is:
\[ (-\infty, 0) \cup (0, \infty) \]