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Math Help - Differentiating

  1. #1
    Junior Member surffan's Avatar
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    Differentiating

    I just want to know if anyone can check to see if i'm solving these questions correctly:

    a) y=xe^2x
    =no idea

    b) f(x)=3^x+x^3
    f'(x)=x(3)^x-1 + 3X^2

    c) y=e^-(2x+5)
    =e^-(2x+5) x ln e x -2

    d) y=e^(3x^2 -5x+7)
    =e^(3x^2 -5x+7) x ln e x 6x-5
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  2. #2
    Senior Member
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    for (a) use. when y=uv
    \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}
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  3. #3
    MHF Contributor

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    Quote Originally Posted by surffan View Post
    I just want to know if anyone can check to see if i'm solving these questions correctly:

    a) y=xe^2x
    =no idea
    Just use the product rule, as arze says.

    b) f(x)=3^x+x^3
    f'(x)=x(3)^x-1 + 3X^2
    No. The formula (x^n)'= n(x^{n-1}) requires the that base be the variable x and the exponent be a constant. Here it is the other way around. The derivative of a^x is (ln a)a^x.


    c) y=e^-(2x+5)
    =e^-(2x+5) x ln e x -2
    PLEASE don't use "x" to mean the variable and to mean times!
    The derivative is, indeed, e^-(2x+5)(-2)(ln e), but ln e= 1. This is just -2e^-(2x+ 5).

    d) y=e^(3x^2 -5x+7)
    =e^(3x^2 -5x+7) x ln e x 6x-5
    Again, don't mix "x" as variable and "times"! Also ln e= 1. This is just (6x-5)e^(3x^2- 5x+ 7).
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  4. #4
    Junior Member surffan's Avatar
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    I gave it a try again and this is what i got

    a) y=xe^2x
    =e^2x +2xe^2x

    b) f(x)= 3^x+x^3
    = 3^xln3+3x^2

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

    d) y= e^(3x^2-5x+7)
    = e^(3x^2-5x+7) (6x-5)

    I reviewed the rules and I hope I got it right this time
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