Use the preceding relation to prove that

$e^{aA+b{A}^{+}}=e^{aA}{e}^{b{A}^{+}}{e}^{-(\frac{1}{2})ab}$

The procedure is the following. Let

$e^{(aA+b{A}^{+})}-=e^{aA}F()$

Differentiation with respect to $$ yields

$(aA+b{A}^{+})e^{(aA+b{A}^{+})}=aA{e}^{aA}F()+e^{aA}\frac{dF}{d}$

that is$,$

$(aA+b{A}^{+})e^{aA}F()=aA{e}^{aA}(F)+e^{aA}\frac{dF}{d}$

Use Problem $17$ to show that

$\frac{dF}{d}=(b{A}^{-1}-ab)F()$

so that

$F()=e^{b{A}^{**}}{e}^{-(\frac{1}{2}){}^{2}ab}$