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anonymous
 one year ago
Help with the Chain Rule for a medal?
anonymous
 one year ago
Help with the Chain Rule for a medal?

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FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0do you have a question in particular?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[3\sqrt[4]{29x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so rerewrite: \[3(29x)^{\frac{1}{4}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0so we have: \[\frac{d}{dx}(3(29x)^{\frac{1}{4}})=\frac{d}{dx}(3(u(x))^{\frac{1}{4}})*\frac{d}{dx}(29x) \] where I defined: \[ u(x)=29x \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so that gives us: \[\frac{d}{dx}(3(29x)^{\frac{1}{4}})=\frac{d}{dx}(3(u(x))^{\frac{1}{4}})*\frac{d}{dx}(29x) = 3*\frac{1}{4}(u(x))^{\frac{3}{4}}*(9)\] The back sub and simplify: \[ ... = \frac{27}{4}(29x)^{\frac{3}{4}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does 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
 one year ago
Best ResponseYou've already chosen the best response.0I'm reading over it again to see if I can better understand it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My 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 (29x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0I understand how you rewrote the function and separated out the composite function...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe I shouldve wrote one of the steps instead as: \[\frac{d}{dx}( 3*(u(x))^{\frac{1}{4}})*\frac{d}{dx}(u(x)) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The first three parts of it make sense, but I'm lost at the 4th

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i.e.: \[f(u(x))=(u(x))^{\frac{1}{4}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well 3 times that anyways

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh! 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
 one year ago
Best ResponseYou've already chosen the best response.0So basically, you separate out the composite functions, then find the derivatives of each, next multiply them together, and finally simplify. Correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correct.... the only trick is correctly identifying the composite function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok challenge question:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is the answer: \[\frac{d}{dx}(a(b(c(d(e(f(x)))))))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can use ' to just denote prime

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and a(x) can stand in for everything inside of it....ah hah gotcha... try this:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{d}{dx}(a(b(x)))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just for shorthand's sake

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I'm going to be honest. I haven't a clue.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its just as before: \[\frac{d}{dx}(a(b(x)))=a'(b(x))*b'(x) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im just using a and b instead of f and u (pardon the last two letters total accident)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what about: \[\frac{d}{dx}(a(b(c(x)))) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well 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
 one year ago
Best ResponseYou've already chosen the best response.0Where 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
 one year ago
Best ResponseYou've already chosen the best response.0Just as before where I said f(u(x)) were the functions f and u both functions of the independent variable x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0I am trying to get you to a very powerful point about the chain rule using a really abstract example.... follow the logic through

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Trust me it isnt as complicated as you think it is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So \[\frac{ d }{ dx } a(b(x))\] would be \[a'(b(x)) * b'(x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0BINGO!!! :D :D and what about: \[\frac{d}{dx}(a(b(c(x)))) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a'(b(c(x))) * b'(c(x))* c'(x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0BINGO :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
 one year ago
Best ResponseYou've already chosen the best response.0I will definitely remember this. I'll take any help I can get with calculus.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whenever 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much for all your help! :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No sweat cya later :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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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