Find stationary point of $y = \frac {e^{2x}} {4 + e^{3x}}$ The curve with equation $y = \frac {e^{2x}} {4 + e^{3x}}$ has one stationary point. Find the exact values of the coordinates of this point.
I got to the point where this is my $\frac {dy} {dx}$:$$\frac{ (4 + e^{3x}) (2e{^{2x}})-e^{2x}(3e^{3x})}{(4+e^{3x})^2} = 0$$
Is this correct to find the stationary points, if it is, how do I get $x $ from this equation?
 A: Developing your numerator, that must be zero, you have the following:
$$
2e^{5x} + 8e^{2x} - 3e^{5x} = 0 \iff e^{5x} = 8e^{2x} \iff 8 = \frac{e^{5x}}{e^{2x}} = e^{3x} \iff \ln 8 = \ln e^{3x} = 3x\ln e = 3x \iff x = \frac{\ln 8}{3} = \frac{\ln2^3}{3} = \frac{3\ln 2}{3} = \ln 2
$$
Notice that you can divide by an exponential because $e^x\neq 0\;\;\forall x\in\mathbb{R}$.
A: $\frac {dy} {dx} = \frac{e^{2x}(3e^{3x}) - (4 + e^{3x}) (2e{^{2x}})}{(4+e^{3x})^2}$
Now,
$$\frac {dy} {dx} = 0$$
$$\frac{e^{2x}(3e^{3x}) - (4 + e^{3x}) (2e{^{2x}})}{(4+e^{3x})^2} = 0$$
$$3e^{5x} - 8e^{2x} - 2e^{5x} = 0$$
$$e^{5x} - 8e^{2x} = 0$$
$$e^{2x}(e^{3x} - 8) = 0$$
Either $$e^{2x} = 0$$
Or $$e^{3x} - 8 = 0$$
Now find $x$ and then put in equation to find $y$.
A: Note the quotient rule:$$\frac{d}{dx} \bigg(\frac{f(x)}{g(x)}\bigg) = \frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{g(x)^2}$$
Then we have $$h(x) = \frac{e^{2x}}{4+e^{3x}},\quad f(x)=e^{2x},\quad g(x) = 4+e^{3x}$$
Then $$h'(x) = \frac{2e^{2x}(4+e^{3x})-e^{2x}\cdot 3e^{3x}}{(4+e^{3x})^2}$$
We set $h'(x) = 0$ to find the stationairy points:
$$ \frac{2e^{2x}(4+e^{3x})-e^{2x}\cdot 3e^{3x}}{(4+e^{3x})^2}=0\implies 8e^{2x}+2e^{5x}-3e^{5x}=0$$
Then $$e^{5x}-8e^{2x}=0\implies e^{2x}(e^{3x}-8)=0 \implies e^{3x}=8$$
Note this is the only solution since $e^{2x}\neq 0$. Then:
$$x=\ln 8\implies x=\frac{1}{3}\ln8 = \ln 2$$ which is the only stationary point. 
A: First we want to know the derivative of $e^{2x}$ using the chain rule and $u = 2x$ and $y= e^u$
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$
$$y' = e^u(2) = 2e^{2x}$$
Applying the quotient rule $(\frac{u}{v})' = \frac{u'v - v'u}{v^2}$
$$y' = \frac{2e^{2x}(4+ e^{3x}) - 3e^{3x}(e^{2x})}{(4+ e^{3x})} = 0$$
This is only zero when the numerator is zero
so
$$2e^{2x}(4+ e^{3x}) - 3e^{3x}(e^{2x}) = 0$$
factoring out the $e^{2x}$ we get
$$8 + 2e^{3x} - 3e^{3x} = 0$$
$$e^{3x} = 8 \implies x = \frac{\ln 8}{3} = \ln 2$$
A: I will use this a chance to publicize a trick I like for solving exponential equations quickly:
Assuming you have already set the numerator to zero
$$
2e^{5x}+8e^{2x}-3e^{5x}=0
$$
set $u=e^x$ and find 
$$
2u^{5}+8u^{2}-3u^{5}=0\implies u=0,u^3=8\implies u=0,2
$$
plugging $u=e^x$ back in, we have 
$$
e^x=0\;\text{or}\:e^x=2
$$
since $e^x>0$ always, $x=\ln 2$.
