Question

Differentiate. F(x) = f ( x f ( x f ( x ) ) ), where f (1)=2, f ' (1)=4, f (2) = 3, f ' (2 ) =5, f ' (3) = 6.
Find F ' (1).

Answer 1

did u differentiate? what are u getting?

Answer 2

Boy this one looks brutal :) lol

\[\large\rm F(x)=f(xf(xf(x)))\]So then,\[\large\rm F'(x)=f'(xf(xf(x)))\cdot \left[xf(xf(x))\right]'\]We'll have product rule for this first chain, yes?

Answer 3

Lemme add some color to the brackets just so we have a better idea of what's going on.\[\rm \color{black}{F'(x)=f'(xf(xf(x)))\cdot \color{orangered}{\left[\color{black}{f(xf(x))+xf'(xf(x))\cdot\color{royalblue}{\left[\color{black}{f(x)+xf'(x)}\right]}}\right]}}\]

Answer 4

For the love of sammy davis... please don't ever use X for multiplication in a problem that involves 17 x's...

seriously... :(

Answer 5

Oh you were just copy pasting maybe XD lol

Answer 6

You can use the asterix *
or \cdot in LaTeX

Answer 7

Tricky little problem :)
We have chain rule into product rule,
into another chain into another product rule.

Are you able to see how those expand out in that way?

Answer 8

From here we just plug in the x-values and it shouldn't simplify down.

Answer 9

\[\frac{ d }{ dx }[xf(xf(x))]\]
How would I do the first expansion

Answer 10

or do I start with f ' ( x f ( x f ( x ) ) first?

Answer 11

We have this:\[\large\rm F(x)=f(xf(xf(x)))\]So we take derivative of outermost f,
and then we'll chain rule in blue,\[\large\rm F'(x)=f'(xf(xf(x)))\cdot \color{royalblue}{\frac{d}{dx}xf(xf(x))}\]

And this thing in blue is what you were asking about,
Product rule:\[\rm \frac{d}{dx}xf(xf(x))\quad= \left(\frac{d}{dx}x\right)f(xf(x))+x\frac{d}{dx}f(xf(x))\]

Answer 12

A bit confusing still? :)

Answer 13

I'm going to attempt the expansion now

Answer 14

Yea, I'm stuck

Answer 15

\[F ' ( x ) = f ' ( x f ( x f ( x ) ) * [ 1 * f ( x f (x ) ) + x \frac{ d }{ dx } f ( x f ( x ) )]\]

Answer 16

Ok looks great so far.

Answer 17

\[x \frac{ d }{ dx }( f ( x f (x) ) = f ' ( x f (x) ) * \frac{ d }{ dx }(x f (x))\]

Answer 18

\[\frac{ d }{ dx }(x f(x) = 1 * f (x) + x f ' (x)\]

Answer 19

Ummm

Answer 20

\[\rm \color{red}{x}\frac{d}{dx}f(xf(x))=\color{red}{x}\left[ f ' ( x f (x) ) * \frac{ d }{ dx }x f (x)\right]\]

Answer 21

Your next chain looks good, just don't forget about that x :)

Answer 22

F ' ( x f ( x f ( x )) * [ f (x f (x) + x f ' (x f ( x ) )] * [ f ( x ) + x f ' (x) ]

Answer 23

@zepdrix I got the full form now, how would I plug the numbers in?

Answer 24

Lemme just fix a couple boo boos,
f ' ( x f ( x f ( x )) * [ f (x f (x)) + x f ' (x f ( x ) ) * [ f ( x ) + x f ' (x) ]]
I moved one of the square brackets,
do you understand why?

This last chain [ f ( x ) + x f ' ( x ) ] should not be multiplying `the whole thing`.
It's only multiplying the x f ' (x f ( x ) )

Answer 25

ahh yes :)

Answer 26

\[\rm F(\color{royalblue}{x})=f ' ( \color{royalblue}{x} f ( \color{royalblue}{x} f ( \color{royalblue}{x} )) * [ f (\color{royalblue}{x} f (\color{royalblue}{x})) + \color{royalblue}{x} f ' (\color{royalblue}{x} f ( \color{royalblue}{x} ) ) * [ f ( \color{royalblue}{x} ) + \color{royalblue}{x} f ' (\color{royalblue}{x}) ]]\]We want to evaluate this function F(x) at x=1,\[\rm F(\color{royalblue}{1})=f ' ( \color{royalblue}{1} f ( \color{royalblue}{1} f ( \color{royalblue}{1} )) * [ f (\color{royalblue}{1} f (\color{royalblue}{1})) + \color{royalblue}{1} f ' (\color{royalblue}{1} f ( \color{royalblue}{1} ) ) * [ f ( \color{royalblue}{1} ) + \color{royalblue}{1} f ' (\color{royalblue}{1}) ]]\]

Answer 27

We can suppress a bunch of 1's that don't mean anything,\[\rm F(\color{royalblue}{1})=f ' ( f ( f ( \color{royalblue}{1} )) * [ f ( f (\color{royalblue}{1})) + f ' ( f ( \color{royalblue}{1} ) ) * [ f ( \color{royalblue}{1} ) + f ' (\color{royalblue}{1}) ]]\]

Answer 28

Then replace all your f(1) with 2's,
and your f '(1) with 4's.

Answer 29

F ' ( 1 ) = f ' ( f ( f ( 1) ) * [ f ( f (1)) + f ' (f ( 1 ) * [ f (1) + f ' (1)] ]
F ' (1) = f ' ( f (2) ) * [ f (2) + f ' (2) * [ f (1) + f '(1)] ]
F ' (1) = f ' ( 3 ) * [ f (2) + f ' (2) * [ f (1) * f ' (1) ] ]

Answer 30

k

Answer 31

Is that form correct?

Answer 32

Pretty crazy to look at

Answer 33

You didn't replace ALL of your f(1) which is kinda strange,
but ya it's a good start so far.

Answer 34

Woops, that last step should be:
F ' (1) = f ' ( 3 ) * [ f (2) + f ' (2) * [ f (1) + f ' (1) ] ]
addition not multiplication in the last brackets.

Answer 35

I'm not getting the same answer as my professor: He gotten 148

\[6 * [ 3 + 5 * [ 2 * 4 ] ] = 384\]

Answer 36

\[\rm F(\color{royalblue}{1})=f ' ( 3) * [ f ( 2) + f ' ( 2 ) * [ f ( \color{royalblue}{1} ) + f ' (\color{royalblue}{1}) ]]\]

Answer 37

\[\rm F(\color{royalblue}{1})=6 * [ 3 + 5 * [ 2 + 4 ]]\]

Answer 38

Which still doesn't match his answer, hmm

Answer 39

Imma work it on paper a sec just to make sure we didn't goof up

Answer 40

Ya 198 seems to be the correct answer.
Maybe teacher made a boo boo somewhere.

Answer 41

What did u have as your final form before putting them into values? The final simplified equation

Answer 42

\[f ' ( 3 ) * [ f (2) + f '( 2 ) . . . \] etc etc

Answer 43

\[\rm \color{black}{F'(x)=f'(xf(xf(x)))\cdot \color{orangered}{\left[\color{black}{f(xf(x))+xf'(xf(x))\cdot\color{royalblue}{\left[\color{black}{f(x)+xf'(x)}\right]}}\right]}}\]\[\rm \color{black}{F'(1)=f'(f(f(1)))\cdot \color{orangered}{\left[\color{black}{f(f(1))+f'(f(1))\cdot\color{royalblue}{\left[\color{black}{f(1)+f'(1)}\right]}}\right]}}\]Replacing f(1) with 2 in every location,\[\rm \color{black}{F'(1)=f'(f(2))\cdot \color{orangered}{\left[\color{black}{f(2)+f'(2)\cdot\color{royalblue}{\left[\color{black}{2+f'(1)}\right]}}\right]}}\]Replacing f(2) with 3 in every location,\[\rm \color{black}{F'(1)=f'(3)\cdot \color{orangered}{\left[\color{black}{3+f'(2)\cdot\color{royalblue}{\left[\color{black}{2+f'(1)}\right]}}\right]}}\]Replacing f'(1) with 4,
f'(2) with 5,\[\rm \color{black}{F'(1)=f'(3)\cdot \color{orangered}{\left[\color{black}{3+5\cdot\color{royalblue}{\left[\color{black}{2+4}\right]}}\right]}}\]

Answer 44

I guess I didn't replace things in the exact same order you did,
so it might look a little different.

Answer 45

Thanks for the help. Appreciate it. I'm going to use yours as a reference to come back to this problem later.

Answer 46

np