Finding the derivative of g(x)

[Biosyn]

Homework Statement

Let f(x) be a continuous and differentiable function on the interval 0 ≤ x ≤ 1, and let g(x)=f(3x). The table below gives values of f'(x), the derivative of f(x). What is the value of g'(0.1)?

http://img845.imageshack.us/img845/442/33806538.jpg

Homework Equations

The Attempt at a Solution

g(0.1) = f(3(0.1))
g(0.1) = f(0.3)
g'(0.1) = f'(0.3)
g'(0.1) = 1.096

Did I do the problem correctly? Thanks!

 

[tazzzdo]

Look correct assuming the picture and problems statement are what you have shown.

 

[shuohg]

I think the answer should be E. g'(x) = 3*f'(3x). So, g'(0.1)=3*f'(0.3)=3*1.096=3.288

 

[tazzzdo]

^ Actually that's correct because of the chain rule (haven't taken calculus in 5 years lol)

g(x) = f(u), where u = 3x so
g'(x) = f'(u)du = f'(3x)*3

 

[Biosyn]

I think the answer should be E. g'(x) = 3*f'(3x). So, g'(0.1)=3*f'(0.3)=3*1.096=3.288

^ Actually that's correct because of the chain rule (haven't taken calculus in 5 years lol)

g(x) = f(u), where u = 3x so
g'(x) = f'(u)du = f'(3x)*3

Thank you guys!
Forgot to use chain rule, thought I could just multiply 3*(0.3)