Key Concepts and Formulas

• Monotonicity of Functions: A function $f(x)$ is increasing if $f'(x) > 0$ for all $x$ in its domain, and decreasing if $f'(x) < 0$ for all $x$ in its domain.
• Quotient Rule: The derivative of $\frac{u(x)}{v(x)}$ is given by $\frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.
• Chain Rule: The derivative of a composite function $f(g(x))$ is given by $f'(g(x)) \cdot g'(x)$.

Step-by-Step Solution

1. State the given function: We are given the function $f(x) = \frac{x}{\sqrt{a^2 + x^2}} - \frac{d - x}{\sqrt{b^2 + (d - x)^2}}$ where $a$, $b$, and $d$ are non-zero real constants and $x \in \mathbb{R}$.

2. Differentiate the first term: Let $u(x) = \frac{x}{\sqrt{a^2 + x^2}}$. We will use the quotient rule to find $u'(x)$. Let $p(x) = x$ and $q(x) = \sqrt{a^2 + x^2}$. Then $p'(x) = 1$ and $q'(x) = \frac{1}{2}(a^2 + x^2)^{-1/2}(2x) = \frac{x}{\sqrt{a^2 + x^2}}$.

   Applying the quotient rule: $u'(x) = \frac{(1)(\sqrt{a^2 + x^2}) - (x)\left(\frac{x}{\sqrt{a^2 + x^2}}\right)}{(\sqrt{a^2 + x^2})^2}$ $u'(x) = \frac{\sqrt{a^2 + x^2} - \frac{x^2}{\sqrt{a^2 + x^2}}}{a^2 + x^2}$ $u'(x) = \frac{\frac{a^2 + x^2 - x^2}{\sqrt{a^2 + x^2}}}{a^2 + x^2}$ $u'(x) = \frac{a^2}{(a^2 + x^2)^{3/2}}$

3. Differentiate the second term: Let $v(x) = \frac{d - x}{\sqrt{b^2 + (d - x)^2}}$. Let $y = d - x$, so $v(x) = \frac{y}{\sqrt{b^2 + y^2}}$. Then $\frac{dy}{dx} = -1$. We can find the derivative with respect to $y$ first, then use the chain rule. $\frac{dv}{dy} = \frac{b^2}{(b^2 + y^2)^{3/2}}$ Now, using the chain rule: $v'(x) = \frac{dv}{dy} \cdot \frac{dy}{dx} = \frac{b^2}{(b^2 + (d - x)^2)^{3/2}} \cdot (-1)$ $v'(x) = -\frac{b^2}{(b^2 + (d - x)^2)^{3/2}}$

4. Combine the derivatives to find $f'(x)$: Since $f(x) = u(x) - v(x)$, then $f'(x) = u'(x) - v'(x)$. $f'(x) = \frac{a^2}{(a^2 + x^2)^{3/2}} - \left(-\frac{b^2}{(b^2 + (d - x)^2)^{3/2}}\right)$ $f'(x) = \frac{a^2}{(a^2 + x^2)^{3/2}} + \frac{b^2}{(b^2 + (d - x)^2)^{3/2}}$

5. Analyze the sign of $f'(x)$: Since $a$ and $b$ are non-zero real constants, $a^2 > 0$ and $b^2 > 0$. Also, $x^2 \ge 0$ and $(d - x)^2 \ge 0$. Therefore, $a^2 + x^2 > 0$ and $b^2 + (d - x)^2 > 0$. This means $(a^2 + x^2)^{3/2} > 0$ and $(b^2 + (d - x)^2)^{3/2} > 0$. Thus, $\frac{a^2}{(a^2 + x^2)^{3/2}} > 0$ and $\frac{b^2}{(b^2 + (d - x)^2)^{3/2}} > 0$. Since $f'(x)$ is the sum of two positive terms, $f'(x) > 0$ for all $x \in \mathbb{R}$.

6. Conclusion: Since $f'(x) > 0$ for all real values of $x$, the function $f(x)$ is an increasing function of $x$.

Common Mistakes & Tips

• Be careful when applying the chain rule, especially with the term $(d-x)$. Remember to multiply by the derivative of the inner function, which is -1 in this case.
• Recognizing patterns in derivatives can save time. For example, the derivative of $\frac{u}{\sqrt{c^2+u^2}}$ with respect to $u$ is $\frac{c^2}{(c^2+u^2)^{3/2}}$.
• Always check the sign of the derivative to determine if the function is increasing or decreasing.

Summary To determine the monotonicity of the given function, we computed its first derivative using the quotient rule and chain rule. We found that the derivative is always positive, indicating that the function is an increasing function of $x$.

Final Answer: The final answer is \boxed{A}, which corresponds to option (A).