How to show that derivative of $\phi(v)$ with respect to $v$ is $\phi'( v)= a(1-\phi^2(v))/2$

How to show that derivative of $\phi(v)$ with respect to $v$ is $$\frac{d \phi}{d v}= \frac{a}{2}(1-\phi^2(v)),$$ where $$\phi(v) = \frac{1-\exp(-av)}{1-\exp(-av)}=\tanh(av/2).$$ What is the value of derivative at the origin? Let's assume that slope parameter $a$ is infinitely big. What kind of equation of $\phi(v)$ you end up?

1 Answer

Well, $\frac{\partial}{\partial v}(\tanh(av/2))=(a/2)$sech$^2(av/2)$, and sech$^2(x)=1-\tanh^2(x)$.