# 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?

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What do you mean by "assume the slope parameter $a$ is infinitely big"? – Alex Becker Feb 19 '12 at 21:41
It might mean: "assume slope parameter a goes to infinity". I quess I'm quite certain of that. – alvoutila Feb 19 '12 at 22:11
So you're asking for the behavior of the solution as $a\to \infty$? – Alex Becker Feb 19 '12 at 22:14
yes. What would be the function $\phi(v)$, if $a \rightarrow \infty$ – alvoutila Feb 19 '12 at 22:47

Well, $\frac{\partial}{\partial v}(\tanh(av/2))=(a/2)$sech$^2(av/2)$, and sech$^2(x)=1-\tanh^2(x)$.
Write out the $\tanh$ function using exponentials and then try to get the above relations. Check the wikipedia article on hyperbolic trigonometric functions. – Sid Raval Feb 19 '12 at 22:24