h = g(f(2x))
dh/dx = dg/df * df/dx * 2
h'(1) = g'(f(2)) * 3 * 2
= g'(3)*6
= 4*6
= 24
f(x) and g(x) are a differentiable function for all reals and h(x) = g[f(2x)]. The table below gives selected values for f(x), g(x), f '(x), and g '(x). Find the value of h '(1). (4 points)
x 1 2 3 4 5 6
f(x) 0 3 2 1 2 0
g(x) 1 3 2 6 5 0
f '(x) 3 2 1 4 0 2
g '(x) 1 5 4 3 2 0
5 answers
i don't know what the right answer is, but 24 was not considered the correct answer
df/dx=2 not 3
noooooooo
16 btw