Answer
See explanations.
Work Step by Step
a. To show $f(x)=\frac{x}{\sqrt {a^2+x^2}}=x(a^2+x^2)^{-1/2}$ increases, we need to test if $f'(x)\gt0$ for all $x$ over the domain. We have $f'(x)=(a^2+x^2)^{-1/2}-\frac{1}{2}x(a^2+x^2)^{-3/2}(2x)=(a^2+x^2)^{-1/2}+x^2(a^2+x^2)^{-3/2}\gt0$ because each individual term is greater than zero.
b. To show $g(x)=\frac{d-x}{\sqrt {b^2+(d-x)^2}}=(d-x)(b^2+(d-x)^2)^{-1/2}$ decreases, we need to test if $g'(x)\lt0$ for all $x$ over the domain. We have $g'(x)=(-1)(b^2+(d-x)^2)^{-1/2}-\frac{1}{2}(d-x)(b^2+(d-x)^2)^{-3/2}(-2(d-x))=-(b^2+(d-x)^2)^{-1/2}+(d-x)^2(b^2+(d-x)^2)^{-3/2}=(b^2+(d-x)^2)^{-3/2}(-b^2-(d-x)^2)+(d-x)^2)=-b^2(b^2+(d-x)^2)^{-3/2}\lt0$ thus $g(x)$ decreases.
c. Comparing this new function with the two functions above, we can write $\frac{dt}{dx}=\frac{f(x)}{c_1}-\frac{g(x)}{c_2}$. Since $c_1, c_2$ are positive constants, the first term will be increasing and the second term (without the minus sign) will be decreasing. Thus, the difference between the two terms will be an increasing function.
