Let $\varphi(x) = \frac{1 - e^{-ax}}{1 + e^{-ax}}$, proof that $\varphi'(x) = \frac{a}{2}(1-\varphi^2(x))$ I am trying to find the required steps to reach that derivative, but I am not finding the right way for that. My current development has the following steps:
$\varphi(x) = \dfrac{1 - e^{-ax}}{1 + e^{-ax}}$, then
$\varphi'(x) = \dfrac{(0-(-ae^{-ax}))(1 + e^{-ax})-(1-e^{-ax})(0-ae^{-ax})}{(1+e^{-ax})^2}$
$\varphi'(x) = \dfrac{ae^{-ax}(1 + e^{-ax})+(1-e^{-ax})ae^{-ax}}{(1+e^{-ax})^2}$
$\varphi'(x) = \dfrac{2ae^{-ax}}{(1+e^{-ax})^2}$.
Now, I am trying to find some way assuming that 
$$\dfrac{1 - e^{-ax}}{1 + e^{-ax}} = \tanh\left(\frac{ax}{2}\right)$$
but without success yet.
 A: Your idea to use hyperbolic functions is good.
First:
$$\varphi(x)=\dfrac{1-e^{-ax}}{1+e^{-ax}}=\dfrac{e^{ax/2}-e^{-ax/2}}{2}\cdot\dfrac{2}{e^{ax/2}+e^{-ax/2}}=\dfrac{\sinh \frac{ax}{2}}{\cosh \frac{ax}{2}}=\tanh\frac{ax}{2}$$
Then
$$\tanh' x=\dfrac{\sinh' x\,\cosh x-\cosh' x\,\sinh x}{\cosh^2x}=\dfrac{\cosh^2x-\sinh^2x}{\cosh^2x}=1-\tanh^2x$$
Hence
$$\varphi'(x)=\dfrac{\mathrm d\left(\tanh \frac{ax}{2}\right)}{\mathrm dx}=\frac{a}{2}\tanh'\frac{ax}{2}=\frac a2(1-\tanh\frac{ax}2)=\frac a2(1-\varphi^2(x))$$
A: You have already shown
$$
\varphi'(x) = \frac{2 a e^{-ax}}{(1+e^{-ax})^2}
$$
On the other hand, the conjectured expression is 
$$
\frac{a}{2}(1-\varphi^2(x)) = \frac{a}{2}(1-\frac{(1-e^{-ax})^2}{(1+e^{-ax})^2}) =  \frac{a}{2}(\frac{4e^{-ax}}{(1+e^{-ax})^2})
$$
So this shows that the conjecture is true!
A: An idea: multiply $\;\phi\;$ by $\;\frac{e^{ax}}{e^{ax}}\;$ , and
$$\phi(x)=\frac{e^{ax}-1}{e^{ax}+1}=1-\frac2{e^{ax}+1}\implies\color{red}{ \phi'(x)=\frac{2ae^{ax}}{(e^{ax}+1)^2}}$$
and OTOH
$$\color{red}{\frac a2\left(1-\phi^2(x)\right)}=\frac a2\left(1-1+\frac4{e^{ax}+1}-\frac4{(e^{ax}+1)^2}\right)=\color{red}{2a\cdot\frac{e^{ax}}{(e^{ax}+1)^2}}$$
and both expressions are identical. Q.E. D..
A: If you see the term $e^{-ax}$ as a function $y$, then it before your eyes will be simple that's $$\varphi'(x) = \frac{2ay}{(1+y)^2} = \frac{a}{2}\frac{(1+y)^2 - (1-y)^2}{(1+y)^2} = \frac{a}{2}(1-\varphi^2(x)).$$
A: You can approach the problem through basic differential equation theory by going "backwards". Suppose that you have the separable ODE 
$$
\tag{$\star$}
\frac{\varphi'}{1-\varphi^2}=\frac{a}{2}
$$
with boundary condition $\varphi(0)=0$ given by the form of $\varphi$ (just calculate it in $0$).
By integrating on both sides you have
$$
\int_0^{\varphi(x)} \frac{\mathrm dy}{1-y^2} = \frac{a}{2}x
$$
so calculating the integral on the left yields
$$
\text{arctanh}(\varphi(x))=\frac{a}{2}x,
$$
which by using the definition of $\tanh$
$$
\tanh(x)=\frac{1-e^{-2x}}{1+e^{-2x}},
$$
becomes
$$
\varphi(x)=\frac{1-e^{-ax}}{1+e^{-ax}}
$$
This shows that if we take $\varphi$ as given by the problem, then $\varphi$ satisfies $(\star)$.
