Differentiate the Function: $y=\frac{ae^x+b}{ce^x+d}$ $y=\frac{ae^x+b}{ce^x+d}$
$\frac{(ce^x+d)\cdot [ae^x+b]'-[(ae^x+b)\cdot[ce^x+d]'}{(ce^x+d)^2}$
numerator only shown (') indicates find the derivative
$(ce^x+d)\cdot(a[e^x]'+(e^x)[a]')+1)-[(ae^x+b)\cdot (c[e^x]'+e^x[c]')+1]$
$(ce^x+d)\cdot(ae^x+(e^x))+1)-[(ae^x+b)\cdot (ce^x+e^x)+1]$
$\frac{(ace^x)^2+de^x+dae^x+de^x+1-[(ace^x)^2(ae^x)^2(bce^x)(be^x)+1]}{(ce^x+d)^2}$
Am I doing this problem correctly? I got to this point and was confused as to how to simplify. Can someone help me?

Another method would be to use logarithmic differentiation. Your function $$y=\frac{ae^x+b}{ce^x+d} $$ is identical to $$\ln(y)=\ln\bigg(\frac{ae^x+b}{ce^x+d}\bigg)=\ln(ae^x+b)-\ln(ce^x+d). $$ We find the differentials for both variables $x$ and $y$. $$\frac{1}{y}dy=\bigg(\frac{ae^x}{ae^x+b}-\frac{ce^x}{ce^x+d}\bigg)dx.$$ We solve for $\frac{dy}{dx}$ and find $$\frac{dy}{dx}=y\bigg(\frac{ae^x}{ae^x+b}-\frac{ce^x}{ce^x+d}\bigg). $$ Substituting the initial equation for $y$ yields a solution of $$\frac{dy}{dx}=\frac{ae^x+b}{ce^x+d}\bigg(\frac{ae^x}{ae^x+b}-\frac{ce^x}{ce^x+d}\bigg). $$ We simplify then distribute through $$\begin{align}\frac{dy}{dx}&=\frac{ae^x+b}{ce^x+d}\bigg(\frac{ae^x(ce^x+d)-ce^x(ae^x+b)}{(ae^x+b)(ce^x+d)}\bigg)\\ &= \frac{e^x[ace^x+ad-(ace^x+bc)]}{(ce^x+d)^2}\\ &=\frac{e^x(ad-bc)}{(ce^x+d)^2}.\end{align}$$ This is your answer.
