Originally Posted by

**skeeter** $\displaystyle D_x$ means the derivative w/r to x.

$\displaystyle \frac{d}{dx} \left(\frac{e^{bx}}{e^{ax}}\right) =$

$\displaystyle \frac{e^{ax} \cdot be^{bx} - e^{bx} \cdot ae^{ax}}{e^{2ax}}=$

$\displaystyle \frac{e^{ax+bx}(b - a)}{e^{2ax}} = e^{bx-ax}(b-a)$

$\displaystyle \frac{f'(x)}{g'(x)} = \frac{be^{bx}}{ae^{ax}} = \frac{b}{a}e^{bx-ax}$

$\displaystyle b-a = \frac{b}{a}$

$\displaystyle ab-a^2 = b

$

$\displaystyle ab-b = a^2$

$\displaystyle b(a-1) = a^2$

$\displaystyle b = \frac{a^2}{a-1}$

btw, you will not see a question like this on the AP exam ... too much time-consuming algebraic manipulation.