How to take the derivative of a function?

New to Julia
1

Let’s say I have a function like this:

T = 10^6
η = 10^9*T^(-3/2) 
x = range(-10, 10, step=1) 
t = [0, 1, 2, 4]
B_z(x, t) = @. erf(x/sqrt(4*η*t))

I wanto differentiate B_z with respect to x and assign it to a new function called dBdx(x, t). Is there a way to do this?

2

There are several options. A good starting point is probably

https://github.com/JuliaDiff/ForwardDiff.jl

3 
T = 10^6
η = 10^9*T^(-3/2) 
x = range(-10, 10, step=1) 
t = [0, 1, 2, 4]
B_z(x, t) = @. erf(x/sqrt(4*η*t))
dB_zdx(x, t) = ForwardDiff.derivative(B_z(x, t), x)

Ok, so I have attempted to set up the derivative as shown here. This does not seem to work as I intended though, because I expect dB_zdx(5, 0) = 0 and dB_zdx(0, 0) = infinity but when I run these I have

MethodError: objects of type Float64 are not callable

Do you know what I may be missing here?

4

certainly. you need to provide the derivative method with a function of x, i.e.

dB_zdx(x, t) = ForwardDiff.derivative(x̃ -> B_z(x̃, t), x)
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