Question

F(x) and g(x) are a differentiable function for all reals and h(x) = g[f(5x)]. 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

Answer 1

By the chain rule,


`h(x)=g(f(5x))\implies h'(x)=5g'(f(5x))f'(5x)`

so


`h'(1)=5g'(f(5))f'(5)`

It's kinda hard to tell which values are which in the table, but if I had to guess, you should get


`h'(t)=5g'(f(5))f'(5)=5g'(2)(0)=0`

Answer 2

The value of h'(1) is 0.

Explanation:

It is given that h(x) = g[f(5x)].

Using chain rule, differentiate with respect to x.

`h'(x)=g'[f(5x)]* f'(5x)* 5`

We have to find the value of h'(1).

`h'(1)=g'[f(5(1))]* f'(5(1))* 5`

`h'(1)=g'[f(5)]* f'(5)* 5`

`h'(1)=g'(2)* (0)* 5`
`[\because f(5)=2, f'(5)=0]`

Product of 0 and any real number is 0.

`h'(1)=0`

Therefore the value of h'(1) is 0.