Using the symbolic toolbox, how do you take a derivative with respect to a derivative?

So I have a function F, which is in terms of time, t, distance, x(t), and velocity, dx/dt = diff(x(t), t). x is defined in terms of time using

syms x(t)

And the function F is, for example,

F = 0.5*x + sqrt(diff(x, t))

But now I need the partial derivative of F in terms of dx/dt. I tried using

derivative = diff(F, diff(x, t))

which results in an error. I tried substituting a letter for dx/dt but differentiation still yields an error.

What is the proper way of taking this derivative? Thanks for the help!

[Walter Roberson]

Would functionalDerivative() help?

[Mark Jansen Graham]

Thanks for the suggestion, that also didn't work for me. "symengine" gives me an error, complaining that "The variable 'diff(x(t), t)' is invalid."

I'm using a workaround by replacing the derivative with another variable, then differentiating F(a) with respect to that variable. So:

G = subs(F, diff(x, t), a)

Then

G_derivative = diff(G, a)
derivative = subs(G_derivative, a, diff(x, t))

This seems like an awfully inconvenient way of doing what I thought would be a fairly simple operation...

[Andrew Newell]

Maybe inconvenience is a matter of taste. Your solution is a lot like the procedure for turning a higher order ODE into a system of linear ODEs.

[Mark Jansen Graham]

You mean like turning a system of n 2nd order ODEs into a system of 2n 1st order ODEs? Kind of, I suppose.

I'm trying to generate equations of motion using Lagrangian mechanics, and a few of the steps requires taking the derivative of F = (x, x') with respect to x and x', so solving for dF/dx', as well as dF/dx.

Solving for dF/dx' only requires one substitution, but dF/dx requires two.

First you would need to replace x' with b, else the second substitution would produce x' = a' = 0

H = subs(F, diff(x, t), b)

then you can replace x with a

I = subs(H, x, a)

and finally, take the derivative with respect to a = x(t)

J = diff(I, H, a)

then restore and b into x and x'

K = subs(J, a, x)
L = subs(K, b, diff(x, t))

where L = dF/dx = the partial derivative of F with respect to x = (t).

This would be just a single line if it were hand-written, but in MATLAB, it's a 5-step process.

I suppose it gets the job done, but I can't help but wonder if there's something a bit more efficient built in to MATLAB.

[Andrew Newell]

There is a book on classical mechanics that argues the usual notation for Lagrangian mechanics is ambiguous. They replace it with a functional notation and provide a language called Scheme to implement it. Maple can do something similar using the D operator. But I'm not sure you can do much better in MATLAB.

[Mark Jansen Graham]

I suppose I'll have to make do with this for now, haha. Thanks!