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).

Question

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)

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)

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

x 1 2 3 4 5 6

Answer 1

We need to take the derivative of h, with respect to x. That is, we need to calculate h'(x).

To do this we will need to use the chain rule.

h(x) = g(f(2x)).

So h'(x) = g'(f(2x))*(f(2x))' by chain rule with g as the outer function and f(2x) is the inner function

= g'(f(2x))*f'(2x)*(2x)' by chain rule with f as the outer function and 2x is the inner function

= g'(f(2x))*f'(2x)*(2) because the derivative of 2x is 2

Now that we have found the derivative we can use the table of values to find the answer.

h'(1) = g'(f(2))*f'(2)*(2) by plugging x = 1 into h'(x).

= g'(3)*f'(2)*(2) because f(2) = 3 from the table

= g'(3)*(2)*(2) because f'(2) = 2 from the table

= (4)*(2)*(2) because g'(3) = 4 from the table

= 16