A teriffic waste of time I'd say...
]]>Differentiating it is easy enough though.
log[sub]e[/sub] [f(x)], when differentiated, becomes f'(x)/f(x)
Applying that here gives that d(log[sub]e[/sub] (1+x))/dx = 1/(1+x).
You'd then need to differentiate that again and again until you get as many terms as you need for your series.
]]>thanks, lloyd
find dy/dx where y= log e (1+x) using the mclaurin series.
(e is supposed to be subscript)
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