Question

Determine the function \( f : R^2->R \) satisies

\( \frac{\partial f}{\partial x}(x,y)=2xy \)  and \( \frac{\partial f}{\partial y}(x,y)=x^2+2y \)

Answer 1

solution 

we have \( \frac{\partial f}{\partial x}(x,y)=2xy \)   (1) and \( \frac{\partial f}{\partial y}(x,y)=x^2+2y \)   (2)

From (2) do integral both sides respected to \( y _1 \) , and it became :

\( \int \frac{\partial f}{\partial y}(x,y)=\int(x^2+2y)dy <=>f(x,y)=x^2y+y^2+c(x) \)  (3)

from (3) , do derivative both side respected to \( x_1 \)

\( => \frac{\partial f}{\partial y}(x,y)=2xy+c'(x) \)   (4)

From (1) and (2) , we obtained as following :

\( 2xy=2xy+c'(x)=>c'(x)=0=>c(x)=k, k\in R \)

Then the function f from (3) become \( f(x,y)=x^2y+y^2+k,k\in R \)

verified: We have \( f(x,y)=x^2y+y^2+k,k\in R \)

\( (*)\frac{\partial f}{\partial x}(x,y)=2xy \)              \( (*)\frac{\partial f}{\partial y}(x,y)=x^2+2y \)

answer 

Thus, \( \frac{\partial f}{\partial x}(x,y)=2xy \) and \( \frac{\partial f}{\partial y}(x,y)=x^2+2y \)