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?
How to show that derivative of $\phi(v)$ with respect to $v$ is $\phi'( v)= a(1-\phi^2(v))/2$
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derivatives
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0yes. What would be the function $\phi(v)$, if $a \rightarrow \infty$ – 2012-02-19
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Well, $\frac{\partial}{\partial v}(\tanh(av/2))=(a/2)$sech$^2(av/2)$, and sech$^2(x)=1-\tanh^2(x)$.
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0Write out the $\tanh$ function using exponentials and then try to get the above relations. Check the wikipedia article on hyperbolic trigonometric functions. – 2012-02-19