> 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
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Reply via email to