How to find the resultant derivative when I know the outputs of $f_{x}$ and $f_{y}$? I have this problem:
Find $h'(1)$ where $h(t)=f(t^{2}-4t,5-4t^{3})$, $f$ is a differentiable function and $f_{x}(-3,1)=2$ and $f_{y}(-3,1)=6$.
My first thought was how do I find the resultant derivative of the differentiable $f$.
Some tips, or ideas to solve this...

$$h'(t) = f_x(t^2-4t,5-4t^3)(\frac{\partial x}{\partial t}) + f_y(t^2-4t,5-4t^3)(\frac{\partial y}{\partial t})$$
$$= f_x(t^2-4t,5-4t^3)(2t-4) + f_y(t^2-4t,5-4t^3)(-12t^2)$$
$$\to h'(1) = f_x(1^2-4(1),5-4(1)^3)(2(1)-4) + f_y((1)^2-4(1),5-4(1)^3)(-12(1)^2)$$
$$ = f_x(-3,1)(-2) + f_y(-3,1)(-12)$$
$$ = (2)(-2) + (6)(-12) = -76$$
