differentiate
a/1+a
I just cant remember how to do it
$\displaystyle \displaystyle \frac{d}{da}\left(\frac{a}{1+a}\right) = \frac{(1+a)\frac{d}{da}(a) - a\,\frac{d}{da}(1+a)}{(1+a)^2}$
$\displaystyle \displaystyle = \frac{(1+a)1 - a(1)}{(1+a)^2}$
$\displaystyle \displaystyle = \frac{1+ a - a}{(1 + a)^2}$
$\displaystyle \displaystyle = \frac{1}{(1+a)^2}$.
Yet a third way! $\displaystyle \frac{a}{a+ 1}= a(a+ 1)^{-1}$. Use the "product rule", rather than the "quotient rule", together with the chain rule:
$\displaystyle \left(a(a+1)^{1}\right)'= (a)'(a+ 1)^{1}+ a((a+1)^{-1})'$
$\displaystyle = (1)(a+ 1)^{-1}+ (a)(-1(a+1)^{-2}(1))$
You could stop there or you could write it as
$\displaystyle = \frac{1}{a+ 1}- \frac{a}{(a+1)^2}= \frac{a+ 1}{(a+1)^2}- \frac{a}{(a+1)^2}= \frac{1}{(a+1)^2}$