Official solution
\[f\left( x \right) = {x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }}\]
\[f'\left( x \right) = {{{a^2}} \over {{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} + {{{b^2}} \over {{{\left( {{b^2} + {{\left( {d - x} \right)}^2}} \right)}^{3/2}}}} > 0\forall x \in R\]
f(x) is an increasing function.