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Thread: 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:
    $\displaystyle \displaystyle y = \sin (\cos\,x)$


    $\displaystyle \displaystyle u = \cos\,x$

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

    $\displaystyle \displaystyle v = \sin\,u$

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


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

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

    $\displaystyle \displaystyle = -\sin^{2}x[\cos(x)] + \cos x [\cos(\cos x)]$
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    $\displaystyle \displaystyle \frac{dy}{dx} = v\frac{du}{dx}\,+\,u\frac{dv}{du}$
    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:

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

    $\displaystyle f(x) = \cos(x)$

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

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


    If you like you can let $\displaystyle u = \cos(x)$ giving $\displaystyle y = \sin(u)$ and then using the chain rule: $\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$
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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:

    $\displaystyle y = sin(cos(x))$

    1.$\displaystyle cos(x) = u(x)$

    Now $\displaystyle y$ becomes $\displaystyle y(u(x))$

    2. Differentiate:

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

    $\displaystyle y = cos(cos(x))*-sin(x)$

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

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

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

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

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

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