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Math Help - Derivative Question

  1. #1
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    Derivative Question

    Hi

    The following question i am having problem finishing off.

    1)For the function P(x) = 5x^{1.5} e^{-2x}

    Find the first and second derivative, and note each point where the first and second derivative is zero. What happens at x=0?

    I have found the first and second derivatives, and is the next step assigning x=0. I done this i always get 0.

    \frac{dy}{dx} = -10x^{0.5}e^{-2x}+7.5x^{0.5}e^{-2x}

    \frac{d^2y}{dx^2} = 20x^{0.5}e^{-2x}-5x^{-0.5}e^{-2x}-15x^{0.5}e^{-2x}+3.75x^{0.5}e^{-2x}



    The next part is to sketch a graph of P(x) against x, noting each of the point identified in (a).

    P.s
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi

    The following question i am having problem finishing off.

    1)For the function P(x) = 5x^{1.5} e^{-2x}

    Find the first and second derivative, and note each point where the first and second derivative is zero. What happens at x=0?

    I have found the first and second derivatives, and is the next step assigning x=0. I done this i always get 0.

    \frac{dy}{dx} = -10x^{\color{red}0.5\color{black}}e^{-2x}+7.5x^{0.5}e^{-2x}

    \frac{d^2y}{dx^2} = 20x^{0.5}e^{-2x}-5x^{-0.5}e^{-2x}-15x^{0.5}e^{-2x}+3.75x^{0.5}e^{-2x}



    The next part is to sketch a graph of P(x) against x, noting each of the point identified in (a).

    P.s
    You made a small error with the product rule. The 0.5 I marked in red should be 1.5. I didn't check the second derivative.
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Paymemoney View Post
    Hi

    The following question i am having problem finishing off.

    1)For the function P(x) = 5x^{1.5} e^{-2x}

    Find the first and second derivative, and note each point where the first and second derivative is zero. What happens at x=0?

    I have found the first and second derivatives, and is the next step assigning x=0. I done this i always get 0.

    \frac{dy}{dx} = -10x^{0.5}e^{-2x}+7.5x^{0.5}e^{-2x}

    \frac{d^2y}{dx^2} = 20x^{0.5}e^{-2x}-5x^{-0.5}e^{-2x}-15x^{0.5}e^{-2x}+3.75x^{0.5}e^{-2x}



    The next part is to sketch a graph of P(x) against x, noting each of the point identified in (a).

    P.s
    Your first derivative doesn't look correct. It should be:

    \frac{dy}{dx} = -10 e^{-2x} x^{1.5}+7.5 e^{-2x} x^{0.5}
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