Question

If
F(x) = f(xf(xf(x))),
where
f(1) = 3, f(3) = 7, f '(1) = 3, f '(3) = 4,
and
f '(7) = 5,
find
F '(1).

Answer 1

I don't quite get how you did that @Loser66
D:

We don't have f(f(f(x))).
We have f(x*f(x*f(x))).

Or you were letting g and h equal x f(...) or something?

Answer 2

I wasn't implying an f^3.
I was reading it like this:
\[\Large\rm F(x)=f\left[x\cdot f(x\cdot f(x))\right]\]Unless it was a typo I guess :o
Which would differentiate something like this.\[\Large\rm F(x)=f'\left[x\cdot f(x\cdot f(x))\right]\cdot \color{royalblue}{(x\cdot f(x\cdot f(x)))'}\]And further uhhhh.....

Answer 3

\[\large\rm F(x)=f'\left[x f(x f(x))\right]\cdot \color{royalblue}{\{f(x f(x))+x f'(x f(x)) f(x)+x f'(x))\}}\]Product rule and stuff.... D:
Oh boy.. what a mess..

Sunyyyy, where you at?

Answer 4

i think you are right

Answer 5

but i am not sure how can i deal with the derivative of x(fx)

Answer 6

you plug in at this point

Answer 7

i guess the answer is 155
5*(4(3+3)+7)

Answer 8

Hmm I got 125, maybe I made a boo boo somwhere.
Where is that 4 coming from?
Isn't that supposed to be a 3?

Answer 9

I dunno, I can't make sense of this anymore XD
and the website is tweaking out on me.. not letting me paste

Answer 10

lol i got 59

Answer 11

maybe i made a mistake

Answer 12

well f'(3) is 4

Answer 13

oops forgot some ( ) on mine
found one of my mistakes

Answer 14

ok 155 is what i got too zep

Answer 15

thank you for your guys, the answer 155 is correct

Answer 16

that was probably one of the ugliest chain rule product combinations i have ever seen
requested

Answer 17

yes, but all my homewok are all like this kind of chain rules combiations so sad