f(u) = u/(u^2-1)
df/du = [(u^2-1) - 2u^2]/(u^2-1)^2
= - (u^2+1)/(u^2-1)^2
fog is just f(g(x)) = f(u)
so, (fog)' is just d/dx(f(g(x)) = df/dx
df/dx = df/du * du/dx =
-(u^2+1)/(u^2-1)^2 * (6x+1)
u(0) = 3
f'(3) = -10/8^2 * 1 = -5/32
I don't see that as one of the choices, but I do see that (a) and (d) are both 5/32, so I assume one of them was -5/32
Find the value of (f o g)' at the given value of x.
f(u)= u/u^2-1 u=g(x)=3x^2+x+3 x=0
is it a. 5/32 b. 13/32 c. 1/8 d. 5/32
1 answer
1 answer