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    how to differentiate this:

    f(x)=sin3x/e^x

    thnx
    note: rep+ available the day after exam.
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    Have you tried using the quotient rule?
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    let u= sin3x v=e^x

    du/dx= 3cos3x
    dv/dx = e^x

    f'(x)= [ v(du/dx) - u(dv/dx) ] / v^2
    = [ e^x(3cos3x ) - sin3x (e^x) ] / (e^x)^2
    =[ e^x ( 3cos3x-sin3x) ] / (e^x)^2
    = ( 3cos3x-sin3x) ] / (e^x).
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    (Original post by FZka)
    let u= sin3x v=e^x

    du/dx= 3cos3x
    dv/dx = e^x

    f'(x)= [ v(du/dx) - u(dv/dx) ] / v^2
    = [ e^x(3cos3x ) - sin3x (e^x) ] / (e^x)^2
    =[ e^x ( 3cos3x-sin3x) ] / (e^x)^2
    = ( 3cos3x-sin3x) ] / (e^x).

    aaah i seee thanks and one more thing
    is (e^x)^2=e^2x or e^x^2
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    e^2x
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    (Original post by DLS)
    e^2x
    thank You for clearing that up now im all done! just need to revise solving equation chapter which won't take long then off to bed and wake up early 8am and start refreshing my mind lol
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    (Original post by Remarqable M)
    aaah i seee thanks and one more thing
    is (e^x)^2=e^2x or e^x^2
    (x^n)^m = x^nm
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    wat wud happen in the situation (3x^2)^3 ???
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    (Original post by Knight91)
    wat wud happen in the situation (3x^2)^3 ???
    27x^6
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    (Original post by Remarqable M)
    how to differentiate this:

    f(x)=sin3x/e^x

    thnx
    note: rep+ available the day after exam.
    :innocent:
 
 
 
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Updated: January 22, 2010
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