Question

Suppose f(x, y) is a differentiable function of x and y and let g(r, s) = f (2rs, 8s − 2r). Use

Answer 1

By the chain rule,

$\dfrac{\partial g}{\partial r} = \dfrac{\partial f}{\partial r} \\\\ ~~~~~ = \dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial r} + \dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial r} \\\\ ~~~~~ = 2 s \dfrac{\partial f}{\partial x} - 2 \dfrac{\partial f}{\partial y}$

where $x(r,s)=2rs$ and $y(r,s)=8s-2r$.

If $r=2$ and $s=1$, then

$x = 2\cdot2\cdot1 = 4$

$y=8\cdot1-2\cdot2 = 4$

so that

$g_r(2,1) = 2\cdot1 f_x(4,4) - 2 f_y(4,4) = 2\cdot2-2\cdot3 = \boxed{-2}$