I am able to compute a symbolic derivative in R for the logit function with this statement:
deriv(quote(exp(-9.3 + 0.0146*x)/(1 + exp(-9.3 + 0.0146*x))),"x")
The result is an expression:
expression({
.expr4 <- exp(-9.3 + 0.0146 * x)
.expr5 <- 1 + .expr4
.expr7 <- .expr4 * 0.0146
.value <- .expr4/.expr5
.grad <- array(0, c(length(.value), 1L), list(NULL, c("x")))
.grad[, "x"] <- .expr7/.expr5 - .expr4 * .expr7/.expr5^2
attr(.value, "gradient") <- .grad
.value
})
However, when I try to return the expression in a function, such as like this:
DerivLogit <- function(x){
deriv(expression(exp(-9.3 + 0.0146*x)/(1 + exp(-9.3 + 0.0146*x))),"x")
}
evaluating DerivLogit(x) of course doesn't substitute the parameter x for the variable x in my expression. And so, for example, DerivLogit(1) equals DerivLogit(2), both of which simply return the expression without any parameter substitution.
Is there a way to convert the derivative expression into a function that I can evaluate where the parameter (e.g. x) will substitute so that I can see the numerical outcome for the given value of x? And if so, how do I do so in R?
You can use eval
to evaluate the expression
DerivLogit <- function(x){
eval(deriv(expression(exp(-9.3 + 0.0146*x)/(1 + exp(-9.3 + 0.0146*x))),"x"))
}