# Evaluation of an Error function

How to evaluate df function in the following code sample?

    f[x_,y_]:=Sqrt[x^4-4x^2 y^2]
df[x_,y_,dx_,dy_]:=Sqrt[(D[f[x,y],x]dx)^2+(D[f[x,y],y] dy)^2]
df[1,2,0.2,0.3]


General::ivar: 1 is not a valid variable. >> General::ivar: 2 is not a valid variable. >>

• := says to evaluate the right hand side later when the function is called. Next df[1,2,0.2,0.3] sets x equal to 1 and y equal to 2. Next evaluate the right hand side doing D[f[1,2],1] and you don't know how to differentiate with respect to 1. Same for 2. If you change the := to = in your definition of the df function and you clear all definitions or restart Mathematica then the right hand side is evaluated before df[1,2..2,.3] and the error message goes away. But you must be very careful with this to be certain that the answer is correct. – Bill May 15 at 3:55
• Thank you @Bill – Emad Raslan May 15 at 4:12
• It works for this sample, but it does not twork for more complicated code which contains many functions. – Emad Raslan May 15 at 4:21
• Please read this tutorial on how to differentiate := and =. It will help you in other contexts too. – Roman May 15 at 7:50
• thank you @Roman – Emad Raslan May 16 at 8:42

## 1 Answer

Use Derivative instead of D in your definitions:

df[x_,y_,dx_,dy_] = Sqrt[(Derivative[1,0][f][x,y] dx)^2+(Derivative[0,1][f][x,y] dy)^2];


Then:

df[1, 2, .2, .3]

0. + 0.95219 I

• I suppose that df[x_,y_,dx_,dy_]=Sqrt[(D[f[x,y],x]dx)^2+(D[f[x,y],y] dy)^2] would work – Claude Leibovici May 15 at 10:01
• Yes, the same @ClaudeLeibovici – Emad Raslan May 16 at 8:44
• many thanks for you @Carl Woll – Emad Raslan May 17 at 21:33