Can I expect ForwardDiff to give the same performance in this case?

General Usage Performance
4

I was able to get similar performance by using const and a package called FastPow This is pretty awesome to me that is why I put it in bold :slight_smile:

EDIT: Results of 0.001 ns cannot be trusted as far as I have been told, but included for good measure - exciting to see how FastPow macro makes it go from 6 ns to that for one calculation.

Benchmarks:

# Manual Function Derivative with FastPow
 @btime WendlandDerivative(u) setup=(u=1.1);
  0.001 ns (0 allocations: 0 bytes)
# Automatic Function Derivative without FastPow
@btime dfNOF(u) setup=(u=1.1);
  6.799 ns (0 allocations: 0 bytes)
# Automatic Function Derivative with FastPow
@btime df(u) setup=(u=1.1);
  0.001 ns (0 allocations: 0 bytes)

# Testing on Arrays
const pts = rand(1000);
# Manual
 @btime WendlandDerivative.(u) setup=(u=pts);
  667.785 ns (1 allocation: 7.94 KiB)
# Automatic
@btime df.(u) setup=(u=pts);
  623.333 ns (1 allocation: 7.94 KiB)

Code below:

using BenchmarkTools
using ForwardDiff
using FastPow

const aD = 696

@fastpow Wendland(q) = aD*(1-q/2)^4 * (2*q+1)
WendlandNOF(q) = aD*(1-q/2)^4 * (2*q+1)
@fastpow WendlandDerivative(q) = aD*((5/8)*q*(q-2)^3)

const dfNOF = x -> ForwardDiff.derivative(WendlandNOF, x)
const df    = x -> ForwardDiff.derivative(Wendland, x)

@btime WendlandDerivative(u) setup=(u=1.1);

@btime dfNOF(u) setup=(u=1.1);
@btime df(u) setup=(u=1.1);

const pts = rand(1000);

@btime WendlandDerivative.(u) setup=(u=pts);

@btime df.(u) setup=(u=pts);