Using the function $f(x,y)$ we define the function:
$g(s,t)=f(s^{2}+\cos(t),s\cdot e^{t})$
It is given that:
$f_{x}(3,0)=-2,f_{y}(3,0)=-4,f_{x}(10,3)=-1,f_{y}(10,3)=6$
What is:
$g_{s}(0,3)$
It is clear that this question involves the chain rule. I know how to solve it, but I think that the numbers are incorrect and that this question cannot be solved. Am I correct ?
I think there is mistaken switch as the comment above suggests. While one cannot apparently calculate $g_s(0,3)$ from the given data, you can calculate $g_s(3,0)$.
Let $x(s,t) = s^2+cos(t)$ and $y(s,t)=se^t$.
Then $\partial x/\partial s=2s$ and $\partial y/\partial s = e^t$.
When $(s,t)=(3,0)$ we have that $(x,y)=(10,3)$.
By the chain rule:
$\frac{\partial g}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$
From the partials calculated above, $\partial x/\partial s|_{(3,0)}=6$ and $\partial y/\partial s|_{(3,0)}=1$.
Now using the third and fourth values given originally (the first two are "red herrings"):
$g_s(3,0) = f_x(10,3)(6)+f_y(10,3)(1) = -1(6)+6(1)=0$.