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### symbolic differentiation of unknown function

I want to do some formal calculus with a unknown functions for the purpose of solving differential equations

Say F(t) = v(t)*t^2, where v is an unknown differentiable function. Then I would like to declare v as such and be able to get

F.diff(t) = 2tv+t^2*v.diff(t)

It is similar to this question, but the solution does not seem to work anymore, as function() takes only one argument and not 2 as in the description; ask.sagemath.org/question/8822/is-there-any-way-to-define-an-as-yet-unknown-function/

Anyone know what the syntax is in 8.9? Or in 9.0, when that becomes available?

 2 None slelievre 17654 ●22 ●160 ●348 http://carva.org/samue...

### symbolic differentiation of unknown function

I want to do some formal calculus with a unknown functions functions for the purpose of solving differential equationsequations.

Say F(t) = v(t)*t^2, v(t)*t^2, where v v is an unknown differentiable function. function.

Then I would like to declare v v as such and be able to get

F.diff(t) = 2tv+t^2*v.diff(t)2*t*v+t^2*v.diff(t)

It is similar to this question, Ask Sage question 8822 but the solution does not seem to work anymore, as function() takes function() takes only one argument and not 2 as in the description; ask.sagemath.org/question/8822/is-there-any-way-to-define-an-as-yet-unknown-function/description.

Anyone know what the syntax is in 8.9? Or in 9.0, when that becomes available?

 3 None slelievre 17654 ●22 ●160 ●348 http://carva.org/samue...

### symbolic differentiation of unknown function

I want to do some formal calculus with unknown functions for the purpose of solving differential equations.

Say F(t) = v(t)*t^2, where v is an unknown differentiable function.

Then I would like to declare v as such and be able to get

F.diff(t) = 2*t*v+t^2*v.diff(t)

It is similar to Ask Sage question 8822 but the solution does not seem to work anymore, as function() takes only one argument and not 2 as in the description.

Anyone know what the syntax is in 8.9? Or in 9.0, when that becomes available?