Find $F_u(1, 1)$ and $F_{u,v}(1, 1)$.

Question

Need help on this.. Suppose that $F(u, v) = f(x(u, v), y(u, v))$, where $f$ is a function satisfying \begin{cases}f(1, 2) = 3\\f_x(1, 2) = 1\\f_y(1, 2) = −2\\f_{x,x}(1, 2) = 3\\f_{x,y}(1, 2) = 2\\f_{y,y}(1, 2) = 0\end{cases}

Suppose further that $x(u, v) = u + v − 1$ and $y(u, v) = 3uv − 1$. Find $F_u(1, 1)$ and $F_{u,v}(1, 1)$.

I found $F_u(1,1)$ but cannot figure out $F_{u,v}(1 , 1)$. The answer sheet says $F_{u,v}(1 , 1)$ is 9 but I keep turning up 6.

Answer 1

By the chain rule, $$F_u = f_x x_u + f_y y_u $$

Then, using the product rule,

\begin{align} F_{u,v} &= (f_x x_u)_v + (f_y y_u)_v \\ &= (f_x)_v x_u + f_x (x_u)_v + (f_y)_v y_u + f_y (y_u)_v \end{align}

Note that $f_x$ and $f_y$ are still functions of $x$ and $y$. At $u = v = 1$,

\begin{align} (f_x)_v &= f_{xx} x_v + f_{xy} y_v = 3 \cdot 1 + 2 \cdot 3 = 9 \\ x_u &= 1 \\ (x_u)_v &= 0 \\ (f_y)_v &= f_{xy} x_v + f_{yy} y_v = 2 \cdot 1 + 0 = 2 \\ y_u &= 3 \\ f_y &= -2 \\ y_{uv} &= 3 \end{align}

Plugging this all back in gives $F_{uv} = 9 \cdot 1 + 0 + 2 \cdot 3 + (-2) \cdot 3 = 9$