Suppose that $F(u, v) = f(x(u, v), y(u, v))$, where $f$ is a function satisfying $f(1, 2) = 3$, $f_x(1, 2) = 1$ and $f_y(1, 2) = −2$. Suppose further that $x(u, v) = u + v − 1$ and $y(u, v) = 3uv − 1$. Find $F_u(1, 1)$. (Note: likely this is a typo.)
Can someone please explain how to solve this because I really don't know how?
The answer is supposedly $-5$.
Given $F(u, v) = f(x(u, v), y(u, v))$, we want to first find $F_u(u, v)$ in order to evaluate it. We know that $f$ depends on both $x,y$ and each of those depend on $u$.
$$F_u(u, v) = f_u(x(u, v), y(u, v)) = f_x(x,y) x_u(u,v) + f_y(x,y) y_u(u,v)$$
Or with less notation,
$$F_u = f_u = f_x x_u + f_y y_u$$
This should give you a start, though as Thomas said in the comments, you can't hope to find $F_u(1,1)$! $F_u(1,2)$ on the other hand does indeed equal -5.