anonymous
  • anonymous
Help with the Chain Rule for a medal?
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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FireKat97
  • FireKat97
do you have a question in particular?
anonymous
  • anonymous
\[-3\sqrt[4]{2-9x}\]
anonymous
  • anonymous
ok so rerewrite: \[-3(2-9x)^{\frac{1}{4}}\]

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anonymous
  • anonymous
so we have something of the form of f(u(x)) thus the derivative will be: \[ \frac{d}{dx}f(u(x))=f'(u(x))*u'(x)\]
anonymous
  • anonymous
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: \[ u(x)=2-9x \]
anonymous
  • anonymous
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: \[ ... = \frac{27}{4}(2-9x)^{-\frac{3}{4}}\]
anonymous
  • anonymous
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?
anonymous
  • anonymous
I'm reading over it again to see if I can better understand it.
anonymous
  • anonymous
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)
anonymous
  • anonymous
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)
anonymous
  • anonymous
I understand how you rewrote the function and separated out the composite function...
anonymous
  • anonymous
Maybe I shouldve wrote one of the steps instead as: \[\frac{d}{dx}( -3*(u(x))^{\frac{1}{4}})*\frac{d}{dx}(u(x)) \]
anonymous
  • anonymous
The first three parts of it make sense, but I'm lost at the 4th
anonymous
  • anonymous
i.e.: \[f(u(x))=(u(x))^{\frac{1}{4}}\]
anonymous
  • anonymous
well -3 times that anyways
anonymous
  • anonymous
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.
anonymous
  • anonymous
So basically, you separate out the composite functions, then find the derivatives of each, next multiply them together, and finally simplify. Correct?
anonymous
  • anonymous
correct.... the only trick is correctly identifying the composite function
anonymous
  • anonymous
Ok challenge question:
anonymous
  • anonymous
Okay
anonymous
  • anonymous
what is the answer: \[\frac{d}{dx}(a(b(c(d(e(f(x)))))))\]
anonymous
  • anonymous
you can use ' to just denote prime
anonymous
  • anonymous
1?
anonymous
  • anonymous
and a(x) can stand in for everything inside of it....ah hah gotcha... try this:
anonymous
  • anonymous
\[\frac{d}{dx}(a(b(x)))\]
anonymous
  • anonymous
Just a second
anonymous
  • anonymous
i.e. a(x)=a(b(x))
anonymous
  • anonymous
just for shorthand's sake
anonymous
  • anonymous
Yeah, I'm going to be honest. I haven't a clue.
anonymous
  • anonymous
sure you do :D:D
anonymous
  • anonymous
Its just as before: \[\frac{d}{dx}(a(b(x)))=a'(b(x))*b'(x) \]
anonymous
  • anonymous
im just using a and b instead of f and u (pardon the last two letters total accident)
anonymous
  • anonymous
so what about: \[\frac{d}{dx}(a(b(c(x)))) \]
anonymous
  • anonymous
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?
anonymous
  • anonymous
Wut?
anonymous
  • anonymous
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
anonymous
  • anonymous
Just as before where I said f(u(x)) were the functions f and u both functions of the independent variable x
anonymous
  • anonymous
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
anonymous
  • anonymous
I am trying to get you to a very powerful point about the chain rule using a really abstract example.... follow the logic through
anonymous
  • anonymous
Trust me it isnt as complicated as you think it is
anonymous
  • anonymous
So \[\frac{ d }{ dx } a(b(x))\] would be \[a'(b(x)) * b'(x)\]
anonymous
  • anonymous
BINGO!!! :D :D and what about: \[\frac{d}{dx}(a(b(c(x)))) \]
anonymous
  • anonymous
\[a'(b(c(x))) * b'(c(x))* c'(x)\]
anonymous
  • anonymous
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
anonymous
  • anonymous
I will definitely remember this. I'll take any help I can get with calculus.
anonymous
  • anonymous
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
anonymous
  • anonymous
Thank you so much for all your help! :D
anonymous
  • anonymous
No sweat cya later :D
anonymous
  • anonymous
@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?

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