# How can I differentiate a symbolic function?

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Federico Avino am 17 Jun. 2019
Beantwortet: Federico Avino am 17 Jun. 2019
I'm in trouble with a problem and I couldn't find any solution for my specific case. I have this code:
syms l1 l2
f = xp
xp_l1 = diff(f,l1)
xp_l2 = diff(f,l2)
l1 and l2 are the two my independent variables depending on time, previously in the code i elaborate the expression of xp, that is dependent on l1 and l2 by a very complex expression. I need the partial derivatives, so i let matlab elaborate them, and it works. The problem come when i need to differentiate xp_l1 and xp_l2 by time. i tried in different ways but nothing allows me to obtain the differentiation automatically. This works but it is too complex and long
syms t l1(t) l2(t)
f = (5*l2(t))/4 - (5*l1(t))/2 + 5*l1(t)*(((200*l1(t) - 100*l2(t))/(25*(100*(l1(t))^2 - 100*l1(t)*l2(t) + 25*(l2(t))^2 + 4)) + ((200*l1(t) - 100*l2(t))*(- 100*(l1(t))^2 + 100*l1(t)*l2(t) - 25*(l2(t))^2 + 252))/(25*(100*(l1(t))^2 - 100*l1(t)*l2(t) + 25*(l2(t))^2 + 4)^2))/(4*((- 100*(l1(t))^2 + 100*l1(t)*l2 - 25*(l2(t))^2 + 252)/(2500*(l1(t))^2 - 2500*l1(t)*l2(t) + 625*(l2(t))^2 + 100))^(1/2)) + 1/2) - (5*l2(t)*(((200*l1(t) - 100*l2(t))/(25*(100*(l1(t))^2 - 100*l1(t)*l2(t) + 25*(l2(t))^2 + 4)) + ((200*l1(t) - 100*l2(t))*(- 100*(l1(t))^2 + 100*l1(t)*l2(t) - 25*(l2(t))^2 + 252))/(25*(100*(l1(t))^2 - 100*l1(t)*l2(t) + 25*(l2(t))^2 + 4)^2))/(4*((- 100*(l1(t))^2 + 100*l1(t)*l2(t) - 25*(l2(t))^2 + 252)/(2500*(l1(t))^2 - 2500*l1(t)*l2(t) + 625*(l2(t))^2 + 100))^(1/2)) + 1/2))/2 - (5*((- 100*(l1(t))^2 + 100*l1(t)*l2(t) - 25*(l2(t))^2 + 252)/(25*(100*(l1(t))^2 - 100*l1(t)*l2(t) + 25*(l2(t))^2 + 4)))^(1/2))/2
xp_l1_t= diff(f,t) % first derivative of wrt t
I wrote the dependance by t by hands. Are there any solutions to do it automatically?
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### Antworten (2)

Bjorn Gustavsson am 17 Jun. 2019
After doing this
syms t
syms l1(t)
syms l2(t)
f = ... % your rather lengthy expression
I had no problem getting an output from diff:
dfdt = diff(f,t)
...and I run a rather old version of matlab. Do you get an error or just no output? It wasn't necessary to split up the symbolic definitions either.
HTH
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Federico Avino am 17 Jun. 2019
Thank you, but my problem is slightly different. I had no problems with the long expression I wrote, but I'm searching for a less wasting time method to do the task. I would like to calculate the derivative wrt time directly from the first part of code, without defining again the syms and expliciting the very long expression, because i have to repeat this tasks a lot of times with a lot of different functions, and this is very time wasting.
I tried this, that teoretically solves my problem, but it doesn't work. Is there any similar solution?
syms t l1(t) l2(t)
f = xp %xp is from another part of code and has the very long expression you read up and has variables l1 and l2, and not l1(t) and l2(t)
xp_l1 = diff(f,l1)
xp_l2 = diff(f,l2)
xp_l1_t= diff(f,t)
xp_l2_t= diff(f,t)
Il presents two issues: the first is that xp has an expression elaborated by another part of code automatically and the expression xp_l1 = diff(f,l1) gives me an error due to the fact that l1 is a symbolic function itself, and the diff can't be performed (only the first object can be a symbolic function). The second issue is that the derivative wrt time gives me always 0
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