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?
 A: Okay so you have $y(x)= \frac{f(x)}{g(x)}$, I have a hard time remembering the rule for differentiating a quotient so this is what I usually do:
$$y(x)= \frac{f(x)}{g(x)} = f(x)\cdot G(x)$$
where $G(x)=\frac{1}{g(x)}$. Then apply the rule for differentiating a product (which I do remember):
$$\frac {\partial y}{\partial dx} = y' = f'(x)\cdot G(x) + f(x)\cdot G'(x)$$
then:
$$ G'(x) = \frac {\partial G}{\partial x} \frac{1}{g(x)} = \frac {\partial G}{\partial x} g(x)^{-1} = -g(x)^{-2}g'(x)$$.
Inserting into $y'$ yields the quotient rule:
$$ y' = \frac{f'(x)}{g(x)} - \frac {f(x)\cdot g'(x)}{g(x)^2} = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$$
Now we can move on to your specific case:
$$f = ae^x+b$$
$$f' = ae^x$$
$$g = ce^x+d$$
$$g' = ce^x$$
. Inserting into the quotient rule for $y'$:
$$y' = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} = \frac{ae^x(ce^x+d) - (ae^x+b)ce^x}{(ce^x+d)^2}$$
Expand numerator, collect terms and simplify:
$$y' = \frac{(ad-bc)e^x}{(ce^x+d)^2}$$
A: $$\frac{ae^x(ce^x+d)-(ae^x+b)ce^x}{(ce^x+d)^2}=
\frac{(ad-bc)e^x}{(ce^x+d)^2}.$$
A: Here are the steps
$$\frac{d}{dx}\left[\frac{ae^x+b}{ce^x+d}\right]$$
$$=\frac{\left(ce^x+d\right)\frac{d}{dx}\left[ae^x+b\right]-\left(ae^x+b\right)\frac{d}{dx}\left[ce^x+d\right]}{\left(ce^x+d\right)^2}$$
$$=\frac{\left(ce^x+d\right)\left(ae^x\right)-\left(ae^x+b\right)\left(ce^x\right)}{\left(ce^x+d\right)^2}$$
$$=\frac{e^x\left(ace^x+ad-ace^x-bc\right)}{\left(ce^x+d\right)^2}$$
$$=\frac{e^x\left(ad-bc\right)}{\left(ce^x+d\right)^2}$$
A: 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.
A: $\frac{ae^x+b}{ce^x+d}=\frac{a}{c}-\frac{ad-bc}{c}(ce^x+d)^{-1}$ implies $f'(x)=(ad-bc)e^x(ce^x+d)^{-2}$.
