Partial derivative evaluation

[Sergey]

Hello!

Would you, please, point out why I can't evaluate the symbolic expression of of the r(x,0) function while d(0) is possible to evaluate? Those are identical expressions but by some reason Mathcad evaluates correctly only one of them. It is inconvinient to introduce new function such as d(t) to compute r(x,t). I marked equations by the blue colour.

Best regards,

Sergey

[LucMeekes · Accepted solution]

You have a function r defined as:

and define a function d with:

and wonder why:

That is because Prime doesn't look back to the definition of r, to check and see if it was defined with a parameter t or not. It sees that r(x,0) is NOT a function of t, so the result of the (partial) derivative is 0.

Note that:

(I replaced t with z...) which means that fun is essentially the same function as your function d.

In order to get:

be the same as the direct form through the partial derivative, you have to:

Success!
Luc

[Sergey]

Thanks a lot, Luc!