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