Question

Help with the Chain Rule for a medal?

Answer 1

do you have a question in particular?

Answer 2

\[-3\sqrt[4]{2-9x}\]

Answer 3

ok so rerewrite:
\[-3(2-9x)^{\frac{1}{4}}\]

Answer 4

so we have something of the form of f(u(x)) thus the derivative will be:
\[ \frac{d}{dx}f(u(x))=f'(u(x))*u'(x)\]

Answer 5

so we have:
\[\frac{d}{dx}(-3(2-9x)^{\frac{1}{4}})=\frac{d}{dx}(-3(u(x))^{\frac{1}{4}})*\frac{d}{dx}(2-9x) \]
where I defined:
\[ u(x)=2-9x \]

Answer 6

Ok so that gives us:

\[\frac{d}{dx}(-3(2-9x)^{\frac{1}{4}})=\frac{d}{dx}(-3(u(x))^{\frac{1}{4}})*\frac{d}{dx}(2-9x) = -3*\frac{1}{4}(u(x))^{-\frac{3}{4}}*(-9)\]
The back sub and simplify:
\[ ... = \frac{27}{4}(2-9x)^{-\frac{3}{4}}\]

Answer 7

Does my notation make sense? I was trying to make it more explicit, but looking back it may look more confusing than it actually is... did this procedure make sense?

Answer 8

I'm reading over it again to see if I can better understand it.

Answer 9

My goal was for you to identify the composition of functions here... one function was raising something to the 1 quarter power... the other function was (2-9x)

Answer 10

The goal of the chain rule is to look at the outermost function of the composition and apply the simple power rule for derivatives... then I can look at (separately) the interior function and differentiate that (also using the deriv power rule)

Answer 11

I understand how you rewrote the function and separated out the composite function...

Answer 12

Maybe I shouldve wrote one of the steps instead as:
\[\frac{d}{dx}( -3*(u(x))^{\frac{1}{4}})*\frac{d}{dx}(u(x)) \]

Answer 13

The first three parts of it make sense, but I'm lost at the 4th

Answer 14

i.e.:
\[f(u(x))=(u(x))^{\frac{1}{4}}\]

Answer 15

well -3 times that anyways

Answer 16

Oh! Okay, I went back and read it, and now it's making sense.
So on the fourth part, you used the power rule on the outside function. That's where I got lost.

Answer 17

So basically, you separate out the composite functions, then find the derivatives of each, next multiply them together, and finally simplify. Correct?

Answer 18

correct.... the only trick is correctly identifying the composite function

Answer 19

Ok challenge question:

Answer 20

Okay

Answer 21

what is the answer:
\[\frac{d}{dx}(a(b(c(d(e(f(x)))))))\]

Answer 22

you can use ' to just denote prime

Answer 23

1?

Answer 24

and a(x) can stand in for everything inside of it....ah hah gotcha... try this:

Answer 25

\[\frac{d}{dx}(a(b(x)))\]

Answer 26

Just a second

Answer 27

i.e. a(x)=a(b(x))

Answer 28

just for shorthand's sake

Answer 29

Yeah, I'm going to be honest. I haven't a clue.

Answer 30

sure you do :D:D

Answer 31

Its just as before:
\[\frac{d}{dx}(a(b(x)))=a'(b(x))*b'(x) \]

Answer 32

im just using a and b instead of f and u (pardon the last two letters total accident)

Answer 33

so what about:
\[\frac{d}{dx}(a(b(c(x)))) \]

Answer 34

Well if the function is x, you would end up taking the derivative of both a'(b(x) and b'(x). Both of them are the function x which would mean both would have a derivative of 1. Multiplying them together 1 x 1 = 1.
Or am I completely off track?

Answer 35

Wut?

Answer 36

Where did you get that the function was x?... a(x) means the function is a... and a(b(x)) means the function a composed with the function b

Answer 37

Just as before where I said f(u(x)) were the functions f and u both functions of the independent variable x

Answer 38

I am being a bit liberal with my function definitions, but this is good if you are confused believe it or not because if it were easy you wouldn't have learned anything... no pain no gain :D

Answer 39

I am trying to get you to a very powerful point about the chain rule using a really abstract example.... follow the logic through

Answer 40

Trust me it isnt as complicated as you think it is

Answer 41

So \[\frac{ d }{ dx } a(b(x))\] would be \[a'(b(x)) * b'(x)\]

Answer 42

BINGO!!! :D :D and what about:
\[\frac{d}{dx}(a(b(c(x)))) \]

Answer 43

\[a'(b(c(x))) * b'(c(x))* c'(x)\]

Answer 44

BINGO :D :D :D..... Take a moment to take this in.... this is actually a really REALLY useful property about derivatives that you can learn to exploit in a lot of different ways to make your life a LOT easier.... but remember if you don't correctly identify the compositions then it all falls apart

Answer 45

I will definitely remember this. I'll take any help I can get with calculus.

Answer 46

Whenever you get a really complicated function that you don't think you can do in one go... just try and break it up into a couple of functions and exploit the chain rule.... as long as you define your composite functions correctly you should never have issues

Answer 47

Thank you so much for all your help! :D

Answer 48

No sweat cya later :D

Answer 49

@PlasmaFuzer

Sorry to summon you back, but I have one last question. Do you always just fill the "u" back in after you're done?