Let \(\mathrm{f}(x)=\frac{x}{\sqrt{\mathrm{a}^2+x^2}}-\frac{\mathrm{d}-x}{\sqrt{\mathrm{~b}^2+(\mathrm{d}-x)^2}}, x \in \mathbb{R}\) where \(a, b, d\) are non-zero real constants. Then

\(\begin{aligned} & \mathrm{f}(x)=\frac{x}{\sqrt{\mathrm{a}^2+x^2}}-\frac{\mathrm{d}-x}{\sqrt{\mathrm{~b}^2+(\mathrm{d}-x)^2}} \\ \therefore \quad & \mathrm{f}^{\prime}(x)=\frac{\sqrt{\mathrm{a}^2+x^2}-\frac{x \times 2 x}{2 \sqrt{\mathrm{a}^2+x^2}}}{\mathrm{a}^2+x^2}\end{aligned}\)
\(\begin{array}{r}-\frac{(-1) \sqrt{b^2+(d-x)^2}+\frac{2(d-x)^2}{2 \sqrt{b^2+(d-x)^2}}}{\left[b^2+(d-x)^2\right]} \\ =\frac{a^2+x^2-x^2}{\left(a^2+x^2\right) \sqrt{a^2+x^2}}-\frac{-\left[b^2+(d-x)^2\right]+(d-x)^2}{\left[b^2+(d-x)^2\right] \sqrt{b^2+(d-x)^2}}\end{array}\)
\(\begin{aligned}
& =\frac{\mathrm{a}^2}{\left(\mathrm{a}^2+x^2\right)^{\frac{3}{2}}}+\frac{\mathrm{b}^2}{\left[\mathrm{~b}^2+(\mathrm{d}-x)^2\right]^{\frac{3}{2}}} \\
& \gt0 \quad \forall x \in \mathrm{R}
\end{aligned}\)
\(\therefore \quad \mathrm{f}(x)\) is an increasing function of \(x\).