Is the following helpful?
pdd<-deriv(~a+(b-a)/(1+exp((c-t)/d)),"d")
> pdd
expression({
.expr1 <- b - a
.expr2 <- c - t
.expr4 <- exp(.expr2/d)
.expr5 <- 1 + .expr4
.value <- a + .expr1/.expr5
.grad <- array(0, c(length(.value), 1L), list(NULL, c("d")))
.grad[, "d"] <- .expr1 * (.expr4 * (.expr2/d^2))/.expr5^2
attr(.value, "gradient") <- .grad
.value
})
Or perhaps you want it with respect to "t"?
JN
Message: 46
Date: Mon, 19 Oct 2009 14:50:15 +0100
From: "Weber, Sam" <sam.we...@exeter.ac.uk>
Subject: [R] How to get slope estimates from a four parameter logistic
with SSfpl?
To: "r-help@R-project.org" <r-help@r-project.org>
Message-ID:
<5bb78285b60f9b4db2c6a4da8f3e788c12c64b3...@exchmbs04.isad.isadroot.ex.ac.uk>
Content-Type: text/plain
Hi,
I was hoping to get some advice on how to derive estimates of slopes from four
parameter logistic models fit with SSfpl.
I fit the model using:
model<-nls(temp~SSfpl(time,a,b,c,d))
summary(model)
I am interested in the values of the lower and upper asymptotes (parameters a
and b), but also in the gradient of the line at the inflection point (c) which
I assume tells me my rate of increase when it is steepest (?).
However, I cannot work out how to derive a slope estimate. I'm guessing it has
something to do with the scaling parameter d but having searched the internet
for hours I have not made any progress, and it is probably quite simple. Any
help would be hugely appreciated!
All the best
Sam
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