please remind how to do - simple differentiation question

**manalive04** · December 21st 2010, 05:23 AM

differentiate

a/1+a

I just cant remember how to do it

**Prove It** · December 21st 2010, 05:27 AM

$\displaystyle \frac{d}{da}\left(\frac{a}{1+a}\right) = \frac{(1+a)\frac{d}{da}(a) - a\,\frac{d}{da}(1+a)}{(1+a)^2}$

$\displaystyle = \frac{(1+a)1 - a(1)}{(1+a)^2}$

$\displaystyle = \frac{1+ a - a}{(1 + a)^2}$

$\displaystyle = \frac{1}{(1+a)^2}$ .

**BAdhi** · December 21st 2010, 06:24 AM

$f(a) = 1 - \frac{1}{a+1}$

$\frac{d(f(a))}{da} = 0 + \frac{1}{(a+1)^2}$

**HallsofIvy** · December 22nd 2010, 03:21 AM

Yet a third way! $\frac{a}{a+ 1}= a(a+ 1)^{-1}$ . Use the "product rule", rather than the "quotient rule", together with the chain rule:
$\left(a(a+1)^{1}\right)'= (a)'(a+ 1)^{1}+ a((a+1)^{-1})'$
$= (1)(a+ 1)^{-1}+ (a)(-1(a+1)^{-2}(1))$

You could stop there or you could write it as
$= \frac{1}{a+ 1}- \frac{a}{(a+1)^2}= \frac{a+ 1}{(a+1)^2}- \frac{a}{(a+1)^2}= \frac{1}{(a+1)^2}$

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