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

1. FireKat97

do you have a question in particular?

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. 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: $u(x)=2-9x$

6. 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: $... = \frac{27}{4}(2-9x)^{-\frac{3}{4}}$

7. 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?

8. anonymous

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

9. 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)

10. 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)

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

well -3 times that anyways

16. 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.

17. anonymous

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

18. anonymous

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

19. anonymous

Ok challenge question:

20. anonymous

Okay

21. anonymous

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

22. anonymous

you can use ' to just denote prime

23. anonymous

1?

24. anonymous

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

25. anonymous

$\frac{d}{dx}(a(b(x)))$

26. anonymous

Just a second

27. anonymous

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

28. anonymous

just for shorthand's sake

29. anonymous

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

30. anonymous

sure you do :D:D

31. anonymous

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

32. anonymous

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

33. anonymous

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

34. 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?

35. anonymous

Wut?

36. 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

37. anonymous

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

38. 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

39. 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

40. anonymous

Trust me it isnt as complicated as you think it is

41. anonymous

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

42. anonymous

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

43. anonymous

$a'(b(c(x))) * b'(c(x))* c'(x)$

44. 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

45. anonymous

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

46. 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

47. anonymous

Thank you so much for all your help! :D

48. anonymous

No sweat cya later :D

49. 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?