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

Answer:

The value of h'(1) is 0.

Step-by-step explanation:

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

Using chain rule, differentiate with respect to x.

$h'(x)=g'[f(5x)]\times f'(5x)\times 5$

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

$h'(1)=g'[f(5(1))]\times f'(5(1))\times 5$

$h'(1)=g'[f(5)]\times f'(5)\times 5$

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

Product of 0 and any real number is 0.

$h'(1)=0$

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