How to plot graph of functions of two variables in a function in R

Question

I made this derivative script and I wanted to plot the graph of the result of the derivative along with the function. But I am not able to make the graph in the multivariate case f(x,y), the result of the derivative is taken as a value and not as a function. See the example below:

DD7<-function(x,r,contador=1){

dy<- substitute(x)
if(contador<1) {

stop("Grau de derivada menor que 1")
}
if(contador==1) {

der<-D(dy,c(r))

print(der)
x<- y <- seq(-3,3,length=50)
z<- outer(x,y,der)
persp(x,y,z)
}
else {
der2<- DD(D(dy,r),r,contador-1)
print(der2)
x2<- y2 <- seq(-3,3,length=50)
z2<- outer(x,y,der2)
persp(x,y,z2)

}
}

DD7(x*y^2,"y",1)

Error in get(as.character(FUN), mode = "function", envir = envir) : 
object 'der' of mode 'function' was not found

Typo? You define DD7 and call DD8. Not reproducible: we don't have your x or y values. — r2evans
– r2evans, Commented Jul 11, 2022 at 19:42
Sorry, I corrected the typo. x and y are variables of the function. The function (DD7) finds the derivative of the typed function. In this example the derivative of (x*(y^2)) with respect to y will answer: (2*xy). I would like to plot the graph of the derivative (2*xy). — Matheus Nunes
– Matheus Nunes, Commented Jul 11, 2022 at 20:11
The third argument for base::outer needs to be a function, whereas der is a call. I don't know why you need outer there, if you call eval(der) in place of that, you get a vector the length of x and y (which, in this example, is the value from y^2, the derivative I think you're targeting). What makes you believe you need to outer the x/y vectors? That'll produce a vector length(x)*length(y) long, fyi. — r2evans
– r2evans, Commented Jul 11, 2022 at 20:16
eval (der) generates a vector, I don't want a vector, I need to turn (der) into a function for me to graph the function. — Matheus Nunes
– Matheus Nunes, Commented Jul 11, 2022 at 20:35

r2evans · Accepted Answer · 2022-07-11 21:18:43Z

As I said in my comments, outer's third argument needs to be a function. If you use eval(der), you may infer (correctly) that it is evaluating the expression (y^2 in this example) based on the objects found in the calling environment (the function's environment, to be specific). Since x and y are both found, then eval(der) uses those variables, and returns a vector of the appropriate length.

The purpose of outer is to do an outer-join of the values, expanding into length(x)*length(y) combinations. If you look at the function as it is called by outer, its first argument is length(x)*length(y) elements long, all from the original x; the second argument is also that length, from the original y. The order of the values are such that all values from x are paired with all values of y.

Given this what we need to do is have eval(der) operate in that environment.

DD7 <- function(x, r, contador=1) {
  # ...
  z2 <- outer(x, y, function(x, y) eval(der))
  persp(x, y, z2)
  # ...
}

DD7(x*y^2, "y", 1)
# x * (2 * y)

In that example, it might be confusing which x and y are referenced when; it's straight-forward: inside the anon-function passed to outer, we also name them the same names so that the evaluated der will find the relevant expanded values.

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