# Help with some partial diff

1. Oct 22, 2004

### matpo39

hi, im having some trouble with this problem:

A certain function f(x,y) is known to have partial derivatives of the form

(partial)f/(partial)x = 2ycos(2x)+y^3*x^2+g(y)
(partial)f/(partial)y= sin(2x)+x^3*y^2+4x+1

Please note that g is a function of y only. Use the equality of mixed partial derivatives (Clairaut's Theorem) to find the function g up to an arbitrary additive constant. then find all the functions f.

i was able to attempt the first part and i got

g'(y)= 4
and then intagrating that i find that g(y)= 4y+c_1

i then inteagrated the two partial functions and i got

f(x)= ysin(2x)+(1/3)y^3*x^3+4yx+c_2
f(y)= ysin(2x)+(1/3)y^3*x^3+4yx+y+c_3

i have no idea on what i should do next, if any one can help me out, that would be great.

thanks

2. Oct 22, 2004

### diegojco

basically are saying you that the function f(x,y) has a gradient equal to (∂f/∂x,∂f/∂y). The condition for a field to be gradient of some function f is the equality of mixed partial derivatives.