Let $f$ be a function of $x$ and $y$ that $f_x = x+2y$ and $f_y = ax+3y$ where $a$ is a constant. In this case, why and what must $a$ be?
My Thoughts:
I think that going backwards is what I have to do here. I have to find $f$. But I'm not sure how to do this.
On
Hint: Presumably, you're comfortable with the following statements: $$\frac{d}{dx}(bx^2)=2bx\qquad\qquad \frac{d}{dx}(cx)=c$$ Thus, if I say that $\frac{d}{dx}(f)=6x$, you know that $f=3x^2$ (potentially with a constant term added on, e.g., $f=3x^2+17$ also works). Remember that for the purposes of differentiation with respect to $x$, the symbol $y$ is a constant. Thus, because $f_x=x+2y$, we must have that $$f=\tfrac{1}{2}x^2+2xy+(\textsf{some expression only involving }y\textsf{ and constants})$$ I'll leave it to you to finish the work though.
As $$f_x=x+2y$$ we must have $$ f=\frac{x^2}{2}+2xy+g(y), $$ where $g(y)$ is a function solely depending on $y$. Thus $$ f_y=2x+g'(y). $$ Since there will be no $x$ appearing in $g'(y)$, we must have $a=2$.