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

**URL:** https://discourse.julialang.org/t/can-i-expect-forwarddiff-to-give-the-same-performance-in-this-case/67894
**Category:** Performance
**Tags:** forwarddiff
**Created:** [September 8, 2021, 7:50pm UTC](https://discourse.julialang.org/t/can-i-expect-forwarddiff-to-give-the-same-performance-in-this-case/67894 "2021-09-08T19:50:22Z")
**Posts on this page:** 1
**Showing post:** 4

<div class="post-metadata">

### Author: ![Ahmed\_Salih](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/ahmed_salih/32/206579_2.png) [@Ahmed\_Salih](https://discourse.julialang.org/u/Ahmed_Salih)
#### Post date: [September 8, 2021, 8:20pm UTC](https://discourse.julialang.org/t/can-i-expect-forwarddiff-to-give-the-same-performance-in-this-case/67894/4 "2021-09-08T20:20:09Z")

</div>

**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 🙂

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:

```julia
# 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:

```julia
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);

```

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