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Math Help - Derivative

  1. #1
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    Derivative

    Hi,

    I'm not arriving at the correct answer, kindly point out my mistake(s).

    Derivative of:
    \displaystyle y = \sin (\cos\,x)


    \displaystyle u = \cos\,x

    \displaystyle \frac{du}{dx} = -\sin\,x

    \displaystyle v = \sin\,u

    \displaystyle \frac{dv}{du} = \cos\,u


    \displaystyle \frac{dy}{dx} = v\frac{du}{dx}\,+\,u\frac{dv}{du}

    \displaystyle = \sin\,u(-\sin\,x) + \cos\,x(\cos\,u)

    \displaystyle = -\sin^{2}x[\cos(x)] + \cos x [\cos(\cos x)]
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    \displaystyle \frac{dy}{dx} = v\frac{du}{dx}\,+\,u\frac{dv}{du}
    That's the product rule, and what you have isn't a product.
    What you've is a composite function, so use the chain rule:

    (f \circ g)'(x) = f'(g(x))g'(x) or \displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.
    Last edited by TheCoffeeMachine; December 4th 2010 at 03:16 PM.
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  3. #3
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    This problem requires the chain rule:

    f(x) = \cos(x)

    g(x) = \sin[f(x)]

    Hence y^{\prime} = g'(x) \cdot f'(x)


    If you like you can let u = \cos(x) giving y = \sin(u) and then using the chain rule: \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}
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  4. #4
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    Okay, just use the chain rule. If you're not familiar with the chain rule, read this:
    Chain rule - Wikipedia, the free encyclopedia

    Anyway, here is the math:

    y = sin(cos(x))

    1.  cos(x) = u(x)

    Now y becomes  y(u(x))

    2. Differentiate:

    \frac {dy}{dx} = \frac {dy}{du} * \frac {du}{dx}

     y = cos(cos(x))*-sin(x)

    What you did is that you used the product rule. Only use the product rule when you have something like this:  y(x) = p(x)q(x) Here, p is a function of x, and q is a function of x as well.

    Your question is however:  y(x) = p(q(x)) It means that p is a function of q(x), and q is a function of x
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  5. #5
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    u = \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \sin u

    u^{'} = -\sin\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y^{'} = \cos u

    (u^{'})(y^{'}) = -\sin x(\cos(\cos x))
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Hellbent View Post
    u = \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \sin u

    u^{'} = -\sin\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y^{'} = \cos u

    (u^{'})(y^{'}) = -\sin x(\cos(\cos x))
    correct!
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