If y=u^3+u^2-1, where u=1/(1-x), determine dy/dx at x=2
y = u^3 + u^2 - 1
dy/du = 3u^2 + 2u
u = 1/(1-x) = (1-x)^-1
du/dx = -1(1-x)^-2 (-1)
= 1/(1-x)^2
when x = 2,
u = 1/(1-2) = -1
du/dx = 1/(1-2)^2 = +1
dy/du = 3(-1)^2 + 2(-1) = 1
dy/dx = dy/du * du/dx
= 1(1) = 1
check my arithmetic, I made several arithmetic errors tonight.