Find a function $z=f(x,y)$ whose partial derivatives are as given, or explain why this is impossible.
We have that $ f_x$ = $ 3x^2y^2-2x$, and $f_y$ = $ 2x^3y+6y$. where $ f_z$ denotes the partial derivative of the function $ f$ with respect to some variable $ z$.
I believe that given the partial derivatives, there is not a function $ z=f(x,y)$ whose partial derivatives are as given.
Pf: We will integrate both $f_x$ and $ f_y$ . The integral of $ f_x$ and the integral of $ f_y$ are not equal by calculus. QED.
Am I correct?
Best Answer
Seeing that $\partial_y f_x = \partial_x f_y$, we see that such $f$ exists.
To go about finding this $f$:
If $f_x(x,y)=3x^2y^2-2x$, then we know that $$f(x,y)=x^3y^2-x^2+c(y)+A$$ where $c(y)$ is constant in $x$, i.e $c$ does not depend on $x$, and $A$ does not depend on either $x,y$ i.e constant.
Now, $f_y(x,y)=2x^3y+6y$ so, $$f(x,y)= x^3y^2+3y^2+d(x)+B$$ where $d$ is constant in $y$, i.e $d$ is independent of $y$, and $B$ does not depend on $x,y$, i.e it is constant.
Now we get $$f(x,y)=x^3y^2-x^2+c(y)+A =x^3y^2+3y^2+d(x)+B$$
We see that we put $$c(y)=3y^2, d(x)=-x^2, A=B $$ then we are done.