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Math Help - derivative e^(5x/2)

  1. #1
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    derivative e^(5x/2)

    Can someone please explain to me how or where do I get this answer from doing this derivative equation...


    y=(x^2-2x+2)e^(5x/2)

    this is where I am stuck:

    y=(x^2-2x+2)e^(5x/2)

    = (x^2-2x+2)(e^(5x/2))(5/2)+(2x-2)(e^5x/2)


    the answer in the book is: (5/2x^2-3x+3)e^(5x/2)

    where do you get the 3x and 3 from?
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  2. #2
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    Re: derivative e^(5x/2)

    Quote Originally Posted by nugen View Post
    Can someone please explain to me how or where do I get this answer from doing this derivative equation...


    y=(x^2-2x+2)e^(5x/2)

    this is where I am stuck:

    y=(x^2-2x+2)e^(5x/2)

    = (x^2-2x+2)(e^(5x/2))(5/2)+(2x-2)(e^5x/2)


    the answer in the book is: (5/2x^2-3x+3)e^(5x/2)

    where do you get the 3x and 3 from?
    $$\large y=\left(x^2-2x+2\right) e^{\dfrac{5x}{2}}$$

    using the product rule for derivatives

    $$\frac{dy}{dx}=\frac{d}{dx} \left(x^2-2x+2\right) \cdot e^{\dfrac{5x}{2}}+\left(x^2-2x+2\right) \cdot \frac{d}{dx}\left(e^{ \dfrac{5x}{2}}\right) =$$

    $$(2x-2)e^{\dfrac{5x}{2}}+\left(x^2-2x+2\right)\frac{5}{2}e^{ \dfrac{5x}{2}} =$$

    $$e^{ \dfrac{5x}{2}}\left(\frac{5}{2}\left(x^2-2x+2\right)+(2x-2)\right)=$$

    $$e^{ \dfrac{5x}{2}}\left(\frac{5}{2}x^2-3x+3\right)$$
    Thanks from nugen
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  3. #3
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    Re: derivative e^(5x/2)

    Multiply everything out. Then combine like terms.
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  4. #4
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    Re: derivative e^(5x/2)

    Your answer is the same as the book's. Pull out the (e^5x/2) term, then simplify this (5/2)(x^2-2x+2) + (2x - 2). You get the same result.
    (e^5x/2)[(5/2)(x^2-2x+2) + (2x - 2)] = (e^5x/2)[(5/2)x^2 -5x + 2x + 5 -2]
    Thanks from nugen
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  5. #5
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    Re: derivative e^(5x/2)

    Quote Originally Posted by nugen View Post
    Can someone please explain to me how or where do I get this answer from doing this derivative equation...
    y=(x^2-2x+2)e^(5x/2)
    this is where I am stuck:
    y=(x^2-2x+2)e^(5x/2)
    = (x^2-2x+2)(e^(5x/2))(5/2)+(2x-2)(e^5x/2)
    the answer in the book is: (5/2x^2-3x+3)e^(5x/2)
    Here is a matter of notation: $e^t=\exp(t)$.

    If $y = \left( {{x^2} - 2x + 2} \right)\exp \left( {2.5x} \right)$ then $y' = \left( {2x - 2} \right)\exp \left( {2.5x} \right) + 2.5\left( {{x^2} - 2x + 2} \right)\exp \left( {2.5x} \right)$
    Thanks from nugen
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