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Math Help - Urgent Help PLEASE ANYONE!!!!

  1. #1
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    Urgent Help PLEASE ANYONE!!!!

    Let y=2e^cos(x)

    A.) Calculate dy/dx and d^2 y/ dx^2

    Show all the work.

    If anyone can help me with this problem it says to show all the work thankyou!!! very urgent help pleaseee!!!!
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  2. #2
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    Krizalid's Avatar
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    Recall that (e^u)'=e^u\cdot u'. (Chain Rule.)
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    Quote Originally Posted by Krizalid View Post
    Recall that (e^u)'=e^u\cdot u'. (Chain Rule.)

    I totally don't know how to do that, if you can show me step by step to find the first and second derivatives of that, would be really helpful.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mmkextreme_1 View Post
    I totally don't know how to do that, if you can show me step by step to find the first and second derivatives of that, would be really helpful.
    Are you saying you have to do this from the limit definition? In terms of the "shortcuts" the only way you can do this problem is using the chain rule. You must have covered that at some point.

    y = 2e^{cos(x)}

    \frac{dy}{dx} = 2 e^{cos(x)} \cdot -sin(x)

    \frac{dy}{dx} = -2~sin(x)~e^{cos(x)}

    Now use the product rule and chain rule again to find the second derivative.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Are you saying you have to do this from the limit definition? In terms of the "shortcuts" the only way you can do this problem is using the chain rule. You must have covered that at some point.

    y = 2e^{cos(x)}

    \frac{dy}{dx} = 2 e^{cos(x)} \cdot -sin(x)

    \frac{dy}{dx} = -2~sin(x)~e^{cos(x)}

    Now use the product rule and chain rule again to find the second derivative.

    -Dan

    hey when you do the second derivative did you get

    d^2 y/ dx^2 =[(2e^cos(x)) * sin^2 x]+ [(2e^cos(x)) * (-sin(x)) * -(-cos(x))]
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mmkextreme_1 View Post
    hey when you do the second derivative did you get

    d^2 y/ dx^2 =[(2e^cos(x)) * sin^2 x]+ [(2e^cos(x)) * (-sin(x)) * -(-cos(x))]
    The product rule is, given h(x) = f(x) \cdot g(x) then h^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)

    The first term looks good, but the second term has some problems.
    \frac{dy}{dx} = -2~sin(x)~e^{cos(x)}

    Let f(x) = sin(x) and g(x) = e^{cos(x)} Then f^{\prime}(x) = cos(x) and g^{\prime}(x) = -sin(x)~e^{cos(x)}. Now put these together.

    -Dan
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    I don't get it can you tell me what you got?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    The product rule is, given h(x) = f(x) \cdot g(x) then h^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)

    The first term looks good, but the second term has some problems.
    \frac{dy}{dx} = -2~sin(x)~e^{cos(x)}

    Let f(x) = sin(x) and g(x) = e^{cos(x)} Then f^{\prime}(x) = cos(x) and g^{\prime}(x) = -sin(x)~e^{cos(x)}. Now put these together.

    -Dan
    Quote Originally Posted by mmkextreme_1 View Post
    I don't get it can you tell me what you got?
    C'mon! I gave you f, g, and their derivatives. All you need to do is plug it into the product rule formula I gave for the derivative of h(x). There is literally almost nothing left to do to get the answer!

    -Dan
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