# Defining all variables as real and positive before differentiation

I'm attempting to differentiate an equation in the form

D[sqrt((2*(((a*b*c+Pi*d*e^2+Pi*f*g^2+h*i*j+Pi*k*l^2)/(a*b*c+Pi*d*e^2+Pi*f*g^2))-1)*m)/(n^2 - o^2)/p), a]


in order to do an error propagation analysis. So I need to differentiate it against a, against b, against c and so on.

All my values will be positive (they reflect physical dimensions of my design) and the value of n will always be greater than o.

Is there a way to define all my variables as real and positive before I differentiate? And to define n greater than o?

The best I've found is $Assumptions = _Symbol [Element] Reals but this only gets me part of the way. ## 2 Answers assume = (And @@ Thread[{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p} > 0]) && n > o;  Note that when you state that a variable is positive then it is automatically also real. And for $Assumptions or Assuming to have an effect, you must use a function that takes the option Assumptions (e.g., Simplify).

Assuming[assume, Element[a, Reals] // Simplify]

(* True *)

expr = Assuming[assume,
D[Sqrt[(2*(((a*b*c + Pi*d*e^2 + Pi*f*g^2 + h*i*j + Pi*k*l^2)/(a*b*c +
Pi*d*e^2 + Pi*f*g^2)) - 1)*m)/(n^2 - o^2)/p], a] // Simplify] This is making the assumption that a is Real and positive, and then it calculates the derivative. But the result is zero.

Assuming[ a \[Element] Reals && a > 0,
D[sqrt ((2*(((abc + Pide^2 + Pifg^2 + hij + Pikl^2)/(abc + Pide^2 +
Pifg^2)) - 1)*m)/(n^2 - o^2)/p), a]]


Hope it helps.

• Thank you. I think your approach might work for me. I modified my equation to give the following....do you think this makes sense? Assuming[{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p} \[Element] Reals && {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p} > 0, D[((2*(((a*b*c + Pi*d*e^2 + Pi*f*g^2 + h*i*j + Pi*k*l^2)/(a*b*c + Pi*d*e^2 + Pi*f*g^2)) - 1)*m)/(n^2 - o^2)/p)^0.5, a]] – DeltaJ Mar 16 '18 at 12:07
• Hi @DeltaJ I think it would do the trick. If not, then you can use the following. a [Element] Reals && a > 0 && b [Element] Reals && b > 0 and so on for the rest of the letters. Hope this helps. Cheers!!! – DiSp0sablE_H3r0 Mar 16 '18 at 12:32
• Appreciate it thank you ;-) – DeltaJ Mar 16 '18 at 13:33