[R] Problems about Derivaties

[enrico.foscolo]

Dear participants to the list,

this is my problem: I want to obtain an expression that represents the second
derivative of one function.
With "deriv3" (package "stats") it is possible to evaluate the second
derivative, but I do not know how I can get the (analytical) expression of this
derivative.

For example: Suppose that I have a function of this form:
f(x,y)=x^3+y^3+(x^2)*(y^2). With "deriv3" I can evaluate the first derivative
and the hessian matrix, as follows:

> d<-deriv3(~x^3+y^3+(x^2)*(y^2),c("x","y"))
> d[[1]]
{
.expr4 <- x^2
.expr5 <- y^2
.expr9 <- 2 * x
.expr15 <- 2 * y
.value <- x^3 + y^3 + .expr4 * .expr5
.grad <- array(0, c(length(.value), 2), list(NULL, c("x",
"y")))
.hessian <- array(0, c(length(.value), 2, 2), list(NULL,
c("x", "y"), c("x", "y")))
.grad[, "x"] <- 3 * .expr4 + .expr9 * .expr5
.hessian[, "x", "x"] <- 3 * .expr9 + 2 * .expr5
.hessian[, "x", "y"] <- .hessian[, "y", "x"] <- .expr9 *
.expr15
.grad[, "y"] <- 3 * .expr5 + .expr4 * .expr15
.hessian[, "y", "y"] <- 3 * .expr15 + .expr4 * 2
attr(.value, "gradient") <- .grad
attr(.value, "hessian") <- .hessian
.value
}
> d[[1]][11]
.hessian[, "x", "y"] <- .hessian[, "y", "x"] <- .expr9 * .expr15()
> typeof(d[[1]][11])
[1] "language"
> is.call(d[[1]][11])
[1] TRUE

Is it possible to extract the formula of the second derivative of this function
from the output of deriv3?

I would like to use this expression as new function, which means that in this
case, I would get the string ".hessian[, "x", "y"] <- .hessian[, "y", "x"] <-
.expr9 * .expr15".

Many thanks for any kind of help in advance,
Enrico



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