# Math Help - reviewing for Finals and STUCK on this! It's not hard, just tricky

1. ## reviewing for Finals and STUCK on this! It's not hard, just tricky

Let F(x) = f(f(x)) and G(x)=(F(x))2 . You also know that f(9)=7, f(7)=3, f′(7)=5, f′(9)=7.

Find F′(9)= find Find G′(9)=

The answers are 35 and 210

The answers are right, but I don't remember how I got this right back when I did it weeks ago? Any explanation will be great! thanks

2. Originally Posted by softballchick
Let F(x) = f(f(x)) and G(x)=(F(x))2 . You also know that f(9)=7, f(7)=3, f′(7)=5, f′(9)=7.

Find F′(9)= find Find G′(9)=

The answers are 35 and 210

The answers are right, but I don't remember how I got this right back when I did it weeks ago? Any explanation will be great! thanks

The chain rule applied:

$F'(9)=[f(f(9))]'=f'(9)f'(f(9))=7 f'(7)=7\cdot 5=35$

$G'(9)=[F(9)^2]'=2F'(9)F(9)=2\cdot 35 f(f(9))=70f(7)=70\cdot 3=210$

Tonio

3. Originally Posted by softballchick
Let F(x) = f(f(x)) and G(x)=(F(x))2 . You also know that f(9)=7, f(7)=3, f′(7)=5, f′(9)=7.

Find F′(9)= find Find G′(9)=

The answers are 35 and 210

The answers are right, but I don't remember how I got this right back when I did it weeks ago? Any explanation will be great! thanks
I'll do the first question leaving the second for you:

To differentiate F(x)=f(f(x)) you have to use the chain-rule:

$F'(x)=f'(f(x)) \cdot f'(x)$

Now plug in x = 9:

$F'(9)=f'(f(9)) \cdot f'(9)$ According to the given values you'll get:

$F'(9)=f'(7) \cdot 7$ which yields finally

$F'(9)= 5 \cdot 7 = 35$

... and now it's your turn