Question

If z = f(x, y), where f is differentiable, and x = g(t) y = h(t) g(1) = 4 h(1) = 7 g' (1) = 5 h' (1) = −3 fx(4, 7) = 4 fy(4, 7) = 7 find dz/dt when t = 1.

Answer 1

$z=f(x,y)=f(g(t),h(t))$

By the chain rule,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial f}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial f}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

but since $z$ is essentially a function of $t$ alone, we can the derivative more succinctly as

$z'(t)=f_x(g(t),h(t))g'(t)+f_y(g(t),h(t))h'(t)$

$z'(1)=f_x(g(1),h(1))g'(1)+f_y(g(1),h(1))h'(1)$
$z'(1)=f_x(4,7)(5)+f_y(4,7)(-3)$
$z'(1)=4(5)+7(-3)$
$z'(1)=-1$