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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

Respuesta :

LammettHash
By the chain rule,

[tex]h(x)=g(f(5x))\implies h'(x)=5g'(f(5x))f'(5x)[/tex]

so

[tex]h'(1)=5g'(f(5))f'(5)[/tex]

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

[tex]h'(t)=5g'(f(5))f'(5)=5g'(2)(0)=0[/tex]

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.

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

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

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

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

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

Product of 0 and any real number is 0.

[tex]h'(1)=0[/tex]

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

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