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Differnentiating e^x

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  • 12-05-2009 2:45pm
    #1
    gimme5minutes
    Registered Users Posts: 387 ✭✭


    Ok so if you differentiate e^x you get e^x. And if you differentiate e^kx, you get ke^kx. So what do you get if you differentiate e^(x-1)?

    It looks like (x-1)e^(x-1), but that seems almost too simple...but is it correct?


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    #2
    Thomas_S_Hunterson
    Closed Accounts Posts: 6,151 ✭✭✭Thomas_S_Hunterson


    Chain rule


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    #3
    gimme5minutes
    Registered Users Posts: 387 ✭✭gimme5minutes


    So e(x-1) would be (x-1)e(x-2).e(x-1)? Does that look alright?


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    #4
    Grudaire
    Registered Users Posts: 3,620 ✭✭✭Grudaire


    gimme5minutes wrote: »
    So e(x-1) would be (x-1)e(x-2).e(x-1)? Does that look alright?

    No, you want to differentiate e^u, where u = x-1
    u = x - 1 => du = dx

    So integral{e^(x-1)dx} = integral{e^u du} = e^u + c = e^(x - 1) + c


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