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Using the Chain Rule

  1. Oct 15, 2006 #1
    Using the "Chain Rule"

    Alright I've been working on some more practice problems and doing fairly well in finding the correct derivative, until I came to this question:

    y = [PLAIN]http://upload.wikimedia.org/math/1/6/0/160a5b4ac79375bd4c5e13c0f3a95f73.pngcos(sin^2x)[/URL] [Broken] (everything is under the square root sign)

    We know that f'(x) = 1/2X^-(1/2) or 1/(2(root)X) and g'(x) = cos(sin^2x) I believe.

    But I'm having troubles trying to imply our chain rule into the problem. Can anyone help me try and interpret this derviative? I just want to say that you guys have been great in helping me understand my mistakes.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 15, 2006 #2
    [tex] y = \sqrt{\cos(\sin^{2}x)}[/tex]. Let [tex] u = \cos(\sin^{2} x) [/tex]. Then [tex] y = \sqrt{u} [/tex]

    Then [tex] \frac{dy}{dx} = \frac{1}{2}u^{-\frac{1}{2}} \frac{du}{dx} [/tex]. What is [tex] \frac{du}{dx} [/tex]?
    Last edited: Oct 15, 2006
  4. Oct 15, 2006 #3
    du/dx would be -sin(sin^2x)*2*sinxcosx right?

    * meaning multiply
  5. Oct 15, 2006 #4
    that is correct.
  6. Oct 15, 2006 #5
    Alright sweet, so now we have to use the chain rule I believe? However, we have to use our du's right, and I'm not sure what steps to do from here.
    Last edited: Oct 15, 2006
  7. Oct 15, 2006 #6
    [tex] \frac{dy}{dx} = \frac{1}{2}u^{-\frac{1}{2}} \frac{du}{dx} [/tex]

    This is the chain rule. multiply [tex] \frac{du}{dx} [/tex] by [tex] \frac{1}{2}u^{-\frac{1}{2}} [/tex].

    We really have [tex] \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} [/tex]
  8. Oct 15, 2006 #7
    Ohhh okay I see what you are saying now. Gotcha amigo! Thanks!
  9. Oct 15, 2006 #8
    One last little question I've got for you for now, thanks a lot everyone, especially you courtrigrad, you've helped out a lot. It asks y= cos(pi)x/ (sin(pi)x+cos(pi)x)

    We know f(x) = cos(pi)x and g(x) = sin(pi)x+cos(pi)x

    y' = dy/dx = g(x)f'(x)-f(x)g'(x)/ g(x)^2

    And we know that the f'(x) = -1 since cos(pi)x = -1x and g'(x) = -1 since sin(pi)x+cos(pi)x= 0x+-1x = -1x

    Subbing into our formula, we get:
    sin(pi)x+cos(pi)x * (-1) - cos(pi)x * (-1)/ (sin(pi)x+cos(pi)x)^2 I believe, however, what's the next step from here in solving this derivative.
  10. Oct 15, 2006 #9
    [tex] y = \frac{\cos(\pi x)}{\sin \pi x + \cos \pi x} [/tex].

    [tex] f'(x) [/tex] does not equal 1. It is equaled to [tex] f'(x) = -\pi\sin \pi x [/tex].

    and [tex] g'(x) = \pi\cos \pi x - \pi\sin \pi x [/tex]
  11. Oct 15, 2006 #10
    Ahhh okay and then you sub those into our formula to get:
    (sin(pi)x+cos(pi)x)(-pi(sin(pi)x) - cos(pi)x*(pi(cos(pi)x-pi(sin(pi)x) / (sin(pi)x+cos(pi)x)^2

    Well basically what you said, but subbing them into our formula, it's hard to type it all out haha. But simplifying this expression proves difficult for me, you basically multiply to get everything out of the brackets, and I've even tried conversing with a few friends but they've been no help in answering, so even if you just show the next step after subbing in all our f(x), g(x), f'(x) and g'(x) values for simplifying this expression, I'm sure I can figure out the rest. You've been a great help though and real patient with me so thank you, if I could I'd give you a cookie haha.
    Last edited: Oct 15, 2006
  12. Oct 15, 2006 #11
    [tex] \frac{(\sin \pi x + \cos \pi x)(-\pi\sin \pi x) - \cos \pi x(\pi\cos \pi x-\pi \sin \pi x)}{(\sin \pi x +\cos \pi x)^{2}} [/tex]

    Expand it out:

    [tex] \frac{-\pi \sin^{2} \pi x - \pi \sin \pi x\cos \pi x - \pi \cos^{2} \pi x + \pi \sin \pi x \cos \pi x}{\sin^{2} \pi x + 2\sin \pi x\cos \pi x + \cos^{2} \pi x} [/tex].

    and use the identity [tex] \sin^{2} x + \cos^{2} x = 1 [/tex]

    you should get: [tex] \frac{-\pi}{2\sin \pi x \cos \pi x + 1}[/tex]
    Last edited: Oct 16, 2006
  13. Oct 16, 2006 #12
    Alrighty here's what I ended up with from that, since the (PI)sin(PI)xcos(PI)x's cancel out, we are left with:
    = -(PI)[sin^2x + cos^2x] / sin^2(PI)x+2sin(PI)xcos(PI)x+cos^2(PI)x
    And using our identity sin^2x + cos^2x = 1, we get,
    -(PI) / sin^2(PI)x+2sin(PI)xcos(PI)x+cos^2(PI)x
    = -(PI) / (sin^2(PI)x + cos^2(PI)x) + 2sin(PI)xcos(PI)x
  14. Oct 16, 2006 #13
    the [tex] \sin^{2}\pi x + \cos^{2} \pi x = 1 [/tex] in the denominator. left out the 1, sorry.
    Last edited: Oct 16, 2006
  15. Oct 16, 2006 #14
    Alright, so it just becomes -(PI)/ 1 + 2sin(PI)xcos(PI)x then, yeah I wasn't too sure whether or not we could use our identity for both the numerator and denominator, alright thanks for your help. :D I just want to say that you've done a fantastic job in helping me answer my problems, even some really silly mistakes you've noticed too along the way with 2/7 rather than 2/3 for that one tangent question, I was going too fast I guess and got carried away, but thanks a bunch.
    Last edited: Oct 16, 2006
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