What is the Derivative of g(x) at x=0 if g(x) = [f(x)]^2?

[lelandsthename]

Homework Statement

If f is differentiable at x=0 and g(x) = [f(x)]^2, f(0) = f'(0) = -1, then g'(0) =

Homework Equations

MC Answers:
(A) -2 (B) -1 (C) 1 (D) 4 (E) 2

The Attempt at a Solution

The only thing I could think of was that if g(x) = (f(x))^2 then g'(0) = (f'(0))^2 and then g'(0) = 1. Does this make sense? I kind of feel like my logic is pseudo math and is giving me an incorrect answer.

 

[rock.freak667]

I would say C) but I can't say if my thinking is correct...it amounts to the same as yours...except for one part

 

[Avodyne]

By the chain rule, g'(x) = 2f(x)f'(x).

 

[HallsofIvy]

Homework Statement

If f is differentiable at x=0 and g(x) = [f(x)]^2, f(0) = f'(0) = -1, then g'(0) =

Homework Equations

MC Answers:
(A) -2 (B) -1 (C) 1 (D) 4 (E) 2

The Attempt at a Solution

The only thing I could think of was that if g(x) = (f(x))^2 then g'(0) = (f'(0))^2 and then g'(0) = 1. Does this make sense? I kind of feel like my logic is pseudo math and is giving me an incorrect answer.

You are right- your logic is pseudo math! The difficulty is that g'(x) is NOT (f'(x))². As Avodyne said, you need to use the chain rule: g(x)= u² and u= f(x). dg/dx= (dg/du)(du/dx).