Finding composite derivatives

[steve snash]

Homework Statement

a) (f ° g)′(−2) = ?
b) (g ° f)′(2) = ?

Homework Equations

f(−2) = −3,
g(−2) = −4,
f(2) = 3,
g(2) = −3,
f ′(−2) = −1,
f ′(−4) = −2,
f ′(2) = 5,
g ′(−2) = 1,
g ′(2) = 2,
g ′(3) = −4.

The Attempt at a Solution

I have no idea how to do it every thing I've tried doesn't work, how do you work it out. Could someone explain the processes so i know how to do any derivative of a composite function

 

[gabbagabbahey]

Just use the chain rule...

$$(f\circ g)(x)=f(g(x))\implies\frac{d}{dx}(f\circ g)(x)=\frac{d}{dx}f(g(x))=\frac{df(g)}{dg}\frac{dg}{dx}=f'(g(x))g'(x)=(f'\circ g)(x)g'(x)$$

 

[lanedance]

i would take the composition of functions to mean:
$$f \circ g(x) = f(g(x))$$

using the chain rule, what is
$$\frac{d}{dx}f(g(x))=?$$

 

[steve snash]

that makes the answer (f o g)'(-2)= f(-2)(g(-2))(g'(-2))
= -1(-4)x(1)
which = 4 which is the wrong answer =(

 

[gabbagabbahey]

that makes the answer (f o g)'(-2)= f(-2)(g(-2))(g'(-2))
= -1(-4)x(1)
which = 4 which is the wrong answer =(

No, $(f'\circ g)(-2)=f'(g(-2))=f'(-4)$