> On Apr 6, 2018, at 3:43 AM, g l <gnuli...@gmx.com> wrote: > >> Sent: Friday, April 06, 2018 at 5:55 AM >> From: "David Winsemius" <dwinsem...@comcast.net> >> >> >> Not correct. You already have `predict`. It is capale of using the `newdata` >> values to do interpolation with the values of the coefficients in the model. >> See: >> >> ?predict >> > > The ยง details did not mention interpolation explicity; thanks. > >> The original question asked for a derivative (i.e. a "gradient"), but so far >> it's not clear that you understand the mathematical definiton of that term. >> We also remain unclear whether this is homework. >> > > The motivation of this post was simple differentiation of a tangent point > (dy/dx) manually, then wondering how to re-think in modern-day computing > terms. Hence the original question about asking the appropriate > functions/syntax to read further ("curiosity"), not the answer (indeed, > "homework"). :) > > Personal curiosity should be considered "homework".

Besides symbolic differentiation, there is also the option of numeric differentiation. Here's an amateurish attempt: myNumDeriv <- function(x){ (exp( predict (graphmodeld, newdata=data.frame(t=x+.0001))) - exp( predict (graphmodeld, newdata=data.frame(t=x) )))/ .0001 } myNumDeriv(c(100, 250, 350)) David Winsemius Alameda, CA, USA 'Any technology distinguishable from magic is insufficiently advanced.' -Gehm's Corollary to Clarke's Third Law ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.