Deriving acceleration from velocity equation

v=sqrt(2*g*H)*tanh((sqrt((2*g*H)/(2*L)))*t)

Answer 1

[Star Strider]

The gradient function could be helpful.

EDIT — (20 May 2023 at 12:46)

If this is symbolic, of course, just take the derivative with respect to t —

syms g H L t 
v = sqrt(2*g*H)*tanh((sqrt((2*g*H)/(2*L)))*t)

v = 

a = diff(v,t)

a = 

.

[Torsten]

My guess is that H is a function of t.

[Star Strider]

Your guess is likely much better than mine.

In that instance, the syms call becomes:

syms g H(t) L t

and the resulting derivative difficult to work with.

However if the result is numeric, gradient would likely still work.