In: Math
f is a twice differentiable function and that its
second partial derivatives are continuous.
Let h(t) =
f (x(t), y(t)) where
x = 2e t and y =
t.
Suppose that fx(2, 0) = 3,
fy(2, 0) = 2,
fxx(2, 0) = 4,
fyy(2, 0) = 3, and
fxy(2, 0) = 1.
h'(t) = fx * 2e t+
fy
find h''(t) when t = 0.
(I don't understand how for the second partial derivative d/dy (fy) = d/dt (x) *fyx + d/dt(y) * fyy
Could you explain this step for me in the solution. Thank you :)