I am having some difficulties solving the following problem:
Let $h: R^{2} \to R^{2}$ a function of class $C^{2}$ and let $f(x, y) = h(x^{2} + y^{2}, x^{3} + y)$.
Determine a non null vector, $v \in R²$, such that $\frac{\partial f} {\partial v} (0, 1) = 0$
Here's how I attempted to do it.
Consider $g(x, y) = (x^{2} + y^{2}, x^{3} + y)$ we have that $f(x, y) = (h \circ g) (x, y)$. Since $h$ and $g$ are of class $C^{2}$ so is $f$ and thus we can conclude that $\frac{\partial f} {\partial v} (0, 1) = Dh({g(0,1)}) Dg(0,1) v$.
Given that g(0, 1) = (1, 1), we have $$ Dh(1,1) = \begin{bmatrix} \frac{\partial h_1} {\partial x} (x,y) & \frac{\partial h_1} {\partial y} (x,y)\\ \frac{\partial h_2} {\partial x} (x,y) & \frac{\partial h_2} {\partial y} (x,y) \end{bmatrix} _ {| (1, 1)} = \begin{bmatrix} \frac{\partial h_1} {\partial x} (1,1) & \frac{\partial h_1} {\partial y} (1,1)\\ \frac{\partial h_2} {\partial x} (1,1) & \frac{\partial h_2} {\partial y} (1,1) \end{bmatrix} $$ and $$ Dg(0,1) = \begin{bmatrix} 2x & 2y\\ 3x^{2} & 1 \end{bmatrix} _ {| (0,1)} = \begin{bmatrix} 0 & 2\\ 0 & 1 \end{bmatrix} $$
Doing the matrix multiplication I end up with $$ \frac{\partial f} {\partial v} (0, 1) = \begin{bmatrix} 0 & 2\frac{\partial h_1} {\partial x} (1,1) + \frac{\partial h_1} {\partial y} (1, 1)\\ 0& 2\frac{\partial h_2} {\partial x} (1,1) + \frac{\partial h_2} {\partial y} (1, 1) \end{bmatrix} v = 0. $$ This is where I get stuck, because the result of this operation will result in a "vector" in R² and the expected result is 0?
Any help is appreciated! I did it like this, and it's likely wrong, because I was having trouble understanding how to apply the chain rule to vector derivatives, I also have no access to the solution or resolution from the source.
Solution:
Thanks to all the replies I managed to understand the problem and this is the entire resolution I managed to come up with. Thanks @TedShifrin for the encouragement!
Given $ v = \begin{bmatrix}a \\b \end{bmatrix} \in R^{2}$ we can solve the equation as follows: $$ \begin{bmatrix} 0 & 2\frac{\partial h_1} {\partial x} (1,1) + \frac{\partial h_1} {\partial y} (1, 1)\\ 0& 2\frac{\partial h_2} {\partial x} (1,1) + \frac{\partial h_2} {\partial y} (1, 1) \end{bmatrix} \begin{bmatrix}a \\b \end{bmatrix} = 0 \iff b \begin{bmatrix} 2\frac{\partial h_1} {\partial x} (1,1) + \frac{\partial h_1} {\partial y} (1, 1)\\2\frac{\partial h_2} {\partial x} (1,1) + \frac{\partial h_2} {\partial y} (1, 1) \end{bmatrix} = 0 \iff b = 0 \lor (2\frac{\partial h_1} {\partial x} (1,1) = -\frac{\partial h_1} {\partial y} (1, 1)) \land (2\frac{\partial h_2} {\partial x} (1,1) = -\frac{\partial h_2} {\partial y} (1, 1)) $$ Hence, all vectors $(a, 0) \in R^{2}, \forall a \in R$ satisfy the criteria.
I am unsure if this is true or possible but given the equation it can be said that:
If $2\frac{\partial h_1} {\partial x} (1,1) = -\frac{\partial h_1} {\partial y} (1, 1) \land 2\frac{\partial h_2} {\partial x} (1,1) = -\frac{\partial h_2} {\partial y} (1, 1)$ then $Df(x, y)$ is the $2x2$ null matrix so all vectors in $R^{2}$ satisfy the criteria.