Derivative a scalar

[bharðarson]

Hello.

I'm trying to derive a formula which is a function of six values. Four of them are just constants but the other two are vectors of three values.

d = constant

L = constant

T = [T1, T2, T3]

T_inf = constant

E = [E1, E2, E3]

R = constant

I want to get three values from the dT_i. Thanks..

[RichardJ]

It's hard to debug a picture. Try using the vectorize operator over the derivative of the function. If that doesn't work, please post the worksheet.

[bharðarson]

Hello Richard.

Thanks for your reply. I couldn't figure this out so here is the file.

I hope someone can help.

BTW. I'm sorry for posting in wrong section. Though I was located in the "Calculus & Derivatives" when posted.

[RichardJ · Accepted solution]

You can't take the derivative w.r.t a subscripted variable, or a variable that is a vector. The easiest solution is to first create a function, and then apply that to the vector variable.

[bharðarson]

I can't open the file but I tried that and it works perfectly.

Thanks.

[RichardJ]

It was created in Prime 3.0. Are you using an earlier version? If so, you should state that in any future questions. Prime 3.0 can't save back to earlier versions, but I would have posted a picture.

[bharðarson]

I'm sorry. I should've mentioned my version. I'm using Prime 2.0.

But deriving the equation before giving the values does the trick.

[VladimirN]

Please find a PDF and XPS version of this worksheet in attachment.

[bharðarson]

Thanks for the pdf. It's not like what my solution was.

But this doesn't work for me... MathCAD says it doesn't recognise the values. 😕

It shows error on d, but it is also on the other values.

[RichardJ]

Delete the inline evaluation on the right hand side of the function definition.

[bharðarson]

Doh, I can't belive I didn't notice that. Thanks!

Now everything works like it should.

Instead of he evaluation on the right I use -> to see the derived equation.