Let $f: \mathbb{R}^2 \to \mathbb{R} $ and $g: \mathbb{R} \to \mathbb{R}^2$.
Suppose $g(0) = (1,2)$, $g'(0) = (-1,4)$ and $(f \circ g)'(0) = -3 $.
Furthermore $\displaystyle \frac{df}{dx}(1,2) = \frac{df}{dy}(1,2)$.
Compute the value of $\displaystyle\frac{df}{dx}(1,2)$
I have used the property that $(f \circ g)'(0) = f' (g(0))* g'(0)$ so I have that $f'(1,2) * (-1,4)$ but I do not know how to compute $f'(1,2)$.
The rules are $$ (f \circ g)(x) = f(g(x)) \\ (f \circ g)'(x) = f'(g(x))(g'(x))\\ $$ we have $$ \frac{df}{dx}(1,2) = \frac{df}{dy}(1,2) = \xi,\qquad g(0) = (1,2), \qquad g'(0) = (-1,4), \qquad (f \circ g)'(0) = -3 $$ \begin{align} -3 &= (f \circ g)'(0) = f'(g(0))(g'(0))\\ &= f'(\begin{pmatrix}1\\2\end{pmatrix})(\begin{pmatrix}-1\\4\end{pmatrix}) \\ &= \begin{pmatrix}\xi & \xi\end{pmatrix}\begin{pmatrix}-1\\4\end{pmatrix}\\ -3&= -\xi+4\xi = 3\xi \implies \xi = -1 = \frac{df}{dx}(1,2) = \frac{df}{dy}(1,2) \end{align} That is $$ f'(1,2) = \begin{pmatrix}-1 & -1\end{pmatrix} $$