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Differentiation of sin and cos

  1. Jun 19, 2004 #1
    i cant work out how to do this question, i have the answer, i just dont know how to get it
    Q. differentiate
    y = sin (cos x)

    A. - cos (cos) sin x

    If any one could show me the solution, it would be greatly appreciated.
     
  2. jcsd
  3. Jun 20, 2004 #2

    jamesrc

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    Gold Member

    Use the chain rule defined something like this:

    [tex] \frac{df(g(x))}{dx} = \frac{df(g)}{dg} \frac{dg(x)}{dx} [/tex]

    Here are the steps:

    [tex] y = \sin \left( \cos x\right ) [/tex]
    [tex] \frac{dy}{dx} = \cos\left(\cos x\right) \frac{d}{dx}\left(\cos x\right) [/tex]
    [tex] \frac{dy}{dx} = \cos\left(\cos x\right) \left(-\sin x\right) [/tex]

    Let me know if anything is still unclear.
     
  4. Jun 20, 2004 #3
    we do the chain rule like this
    sin (cos x) = uv

    where u = sin, v = cos x

    dy/dx = v (du/dx) + u (dv/dx)

    = cos x . cos + sin . -sin x

    Please tell me where i went wrong in this. i know its in my working, thats what i need help with, i get what your saying.
     
  5. Jun 20, 2004 #4
    how stupid of me, i didnt even need to use the chain rule, sorry man, i totally screwed it up, i get how to do it now, thanks
     
  6. Jun 20, 2004 #5
    Yes, you do need to use the chain rule.

    This is the product rule, not the chain rule. sin (cos x) isn't the product of sine and cosine, it's a composite function.
     
  7. Jun 20, 2004 #6
    yeh sorry bout the wrong rule: i just did it like we normally differentiate

    sin (cos x)
    dy/dx = cos (cos x) x (derivative of brackets) which is -sin x
    sorry for such a debate
     
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