R help - Apr 2018 - Obtain gradient at multiple values for exponetial decay model

Readers,

Data set:

t,c
0,100
40,78
80,59
120,38
160,25
200,21
240,16
280,12
320,10
360,9
400,7

graphdata<-read.csv('~/tmp/data.csv')
graphmodeld<-lm(log(graphdata[,2])~graphdata[,1])
graphmodelp<-exp(predict(graphmodeld))
plot(graphdata[,2]~graphdata[,1])
lines(graphdata[,1],graphmodelp)

Please what is the function and syntax to obtain gradient values for the model
curve at various requested values, e.g.:

when graphdata[,1] at values = 100, 250, 350 ?

This smells like homework, which the Posting Guide indicates is off topic.

I am not aware of "the function" that will solve this, but if you know
what a gradient is analytically then you should be able to put together a
solution very similar to the code you already have with the addition of using
the coef function.
-- 
Sent from my phone. Please excuse my brevity.

On April 5, 2018 3:44:03 AM PDT, g l <gnulinux at gmx.com>
wrote:
>Readers,
>
>Data set:
>
>t,c
>0,100
>40,78
>80,59
>120,38
>160,25
>200,21
>240,16
>280,12
>320,10
>360,9
>400,7
>
>graphdata<-read.csv('~/tmp/data.csv')
>graphmodeld<-lm(log(graphdata[,2])~graphdata[,1])
>graphmodelp<-exp(predict(graphmodeld))
>plot(graphdata[,2]~graphdata[,1])
>lines(graphdata[,1],graphmodelp)
>
>Please what is the function and syntax to obtain gradient values for
>the model curve at various requested values, e.g.:
>
>when graphdata[,1] at values = 100, 250, 350 ?
>
>______________________________________________
>R-help at 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.

> Sent: Thursday, April 05, 2018 at 4:40 PM
> From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us>
> 
> the coef function.
> 
For the benefit of other novices, used the following command to read the
documentation:

?coef

Then tried and obtained:
> cvalue100<-coef(graphmodelp~100)
> cvalue100
NULL

Then looked at the model values which of course correspond to original
non-modelled values.

graphmodelp
        1         2         3         4         5         6         7         8 
91.244636 69.457794 52.873083 40.248368 30.638107 23.322525 17.753714 13.514590 
        9        10        11 
10.287658  7.831233  5.961339

This prompted to think that interpolation is required, but the function
'approx' only seems to perform constant interpolation.

Is the correct thinking to find a function to perform interpolation, then
find/write a function to differentiate the model at a specific value of x, to
find gradient at that point?