1. ## Say What?

Suppose that $f$ and $g$ are functions such that:

$f(2)=-1, f'(2)=4, f''(2)=-2, g(2)=-3, g'(2)=2, g''(2)=1$.
Find the value of each of the following at $x=2$:

a) $(2f-3g)'$

b) $(2f-3g)''$

c) $(fg)'$

Would I be right by saying that I need to do something like this:

$(2f-3g)'=2f'-3g'=2(4)-3(2)=2$ ?

Could someone give a small lecture on what's happening here?

Thankx.

2. Originally Posted by VonNemo19
Suppose that $f$ and $g$ are functions such that:

$f(2)=-1, f'(2)=4, f''(2)=-2, g(2)=-3, g'(2)=2, g''(2)=1$.
Find the value of each of the following at $x=2$:

a) $(2f-3g)'$

b) $(2f-3g)''$

c) $(fg)'$

Would I be right by saying that I need to do something like this:

$(2f-3g)'=2f'-3g'=2(4)-3(2)=2$ ? Mr F says: Yes.

Could someone give a small lecture on what's happening here?

Thankx.
You seem to have the idea, so what's to say? Find the required derivatives. Substitute x = 2. Use the given values to evaluate the derivatives.

3. Originally Posted by mr fantastic
You seem to have the idea, so what's to say? Find the required derivatives.
You really are a rascal!

O.K.

So for c) then...

$f'g+g'f=4(-3)+2(-1)=-14$ ?

4. Originally Posted by VonNemo19
You really are a rascal!

O.K.

So for c) then...

$f'g+g'f=4(-3)+2(-1)=-14$ ?
Yes to both.

5. At some point, very early in your Calculus class, you should learn:

(f+ g)'= f'+ g'
(f- g)'= f'- g'
(fg)'= f'g+ fg'
$(f/g)'= (f'g- fg')/g^2$
You need the first three of those here.

6. Originally Posted by HallsofIvy
At some point, very early in your Calculus class, you should learn:

(f+ g)'= f'+ g'
(f- g)'= f'- g'
(fg)'= f'g+ fg'
$(f/g)'= (f'g- fg')/g^2$
You need the first three of those here.
Yeah, I know that. I even understand the proofs behind the elementary differentiation formulas. It's just that, when I initially looked at this problem, I was confused because of the notation used. We were studying the derivatives of inverse functions at the time, and then my techer threw this gem in there. Threw me for a loop.