[R] Partial Derivatives in R

[David Winsemius]

On May 10, 2009, at 10:05 AM, Paul Heinrich Dietrich wrote:

>
> Quick question:
>
> Which function do you use to calculate partial derivatives from a  
> model
> equation?
>
> I've looked at deriv(), but think it gives derivatives, not partial
> derivatives.

Your reading of the help page and the examples differs from mine. It  
says:
" It returns a call for computing the expr and its (partial)  
derivatives, simultaneously."

 > dx2x <- deriv(~ x^2, "x") ; dx2x  # the example on the help page
expression({
     .value <- x^2
     .grad <- array(0, c(length(.value), 1L), list(NULL, c("x")))
     .grad[, "x"] <- 2 * x
     attr(.value, "gradient") <- .grad
     .value
})
 > dyx2.x <- deriv(~ y*x^2, "x") ; dyx2.x
expression({
     .value <- y * x^2
     .grad <- array(0, c(length(.value), 1L), list(NULL, c("x")))
     .grad[, "x"] <- y * (2 * x)
     attr(.value, "gradient") <- .grad
     .value
})
 > dy2x2.x <- deriv(~ y^2*x^2, "x") ; dy2x2.x
expression({
     .expr1 <- y^2
     .value <- .expr1 * x^2
     .grad <- array(0, c(length(.value), 1L), list(NULL, c("x")))
     .grad[, "x"] <- .expr1 * (2 * x)
     attr(.value, "gradient") <- .grad
     .value
})
 > dy2x2.xy <- deriv(~ y^2*x^2, c("x","y")) ; dy2x2.xy
expression({
     .expr1 <- y^2
     .expr2 <- x^2
     .value <- .expr1 * .expr2
     .grad <- array(0, c(length(.value), 2L), list(NULL, c("x",
         "y")))
     .grad[, "x"] <- .expr1 * (2 * x)
     .grad[, "y"] <- 2 * y * .expr2
     attr(.value, "gradient") <- .grad
     .value
})


> Of course my equation isn't this simple, but as an example,
> I'm looking for something that let's you control whether it's a  
> partial or
> not, such as:
>
> somefunction(y~a+bx, with respect to x, partial=TRUE)
>


That appears to be precisely what your are offered with deriv, ....  
just not needing the partial=TRUE

deriv(
David Winsemius, MD
Heritage Laboratories
West Hartford, CT