Let `f(x) = (x)/(sqrt(a^(2) + x^(2)))- (d-x)/(sqrt(b^(2) + (d-x)^(2))), x in R`, where a, b and d are non-zero real constants. Then,

Options

A. f is a decreasing function of x
B. f is neither increasing nor decreasing function of x
C. f' is not a continuous function of x
D. f is an increasing function of x

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x}{\sqrt{a^2 + x^2}} - \frac{d - x}{\sqrt{b^2 + (d - x)^2}} \) and determine whether it is increasing or decreasing.

### Step-by-Step Solution:

1. **Define the Function**:
\[ f(x) = \frac{x}{\sqrt{a^2 + x^2}} - \frac{d - x}{\sqrt{b^2 + (d - x)^2}} \]

2. **Find the Derivative**:
To determine if the function is increasing or decreasing, we need to find the derivative \( f'(x) \).

3. **Use the Quotient Rule**:
We will differentiate each term using the quotient rule. The derivative of \( \frac{u}{v} \) is given by \( \frac{u'v - uv'}{v^2} \).

- For the first term \( \frac{x}{\sqrt{a^2 + x^2}} \):
- Let \( u = x \) and \( v = \sqrt{a^2 + x^2} \).
- Then \( u' = 1 \) and \( v' = \frac{x}{\sqrt{a^2 + x^2}} \).
- Thus,
\[ \frac{d}{dx} \left( \frac{x}{\sqrt{a^2 + x^2}} \right) = \frac{1 \cdot \sqrt{a^2 + x^2} - x \cdot \frac{x}{\sqrt{a^2 + x^2}}}{a^2 + x^2} = \frac{\sqrt{a^2 + x^2} - \frac{x^2}{\sqrt{a^2 + x^2}}}{a^2 + x^2} \]

- For the second term \( \frac{d - x}{\sqrt{b^2 + (d - x)^2}} \):
- Let \( u = d - x \) and \( v = \sqrt{b^2 + (d - x)^2} \).
- Then \( u' = -1 \) and \( v' = \frac{-(d - x)}{\sqrt{b^2 + (d - x)^2}} \).
- Thus,
\[ \frac{d}{dx} \left( \frac{d - x}{\sqrt{b^2 + (d - x)^2}} \right) = \frac{-1 \cdot \sqrt{b^2 + (d - x)^2} - (d - x) \cdot \frac{-(d - x)}{\sqrt{b^2 + (d - x)^2}}}{b^2 + (d - x)^2} \]

4. **Combine the Derivatives**:
Now we combine the derivatives from both terms to find \( f'(x) \).

5. **Simplify the Expression**:
After substituting and simplifying, we will find that:
\[ f'(x) = \frac{a^2}{(a^2 + x^2)^{3/2}} + \frac{b^2}{(b^2 + (d - x)^2)^{3/2}} \]

6. **Analyze the Sign of the Derivative**:
Since both \( a^2 \) and \( b^2 \) are positive constants (as \( a, b, d \) are non-zero), it follows that:
\[ f'(x) > 0 \quad \text{for all } x \in \mathbb{R} \]

7. **Conclusion**:
Since the derivative \( f'(x) \) is positive for all \( x \), the function \( f(x) \) is increasing for all real numbers \( x \).

### Final Answer:
The function \( f(x) \) is increasing for all \( x \in \mathbb{R} \).