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Thread: Derivative – what is wrong in my calculation ?

  1. #1

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    Derivative – what is wrong in my calculation ?

    I started learning calculus. I am in the derivatives stage.

    I am following the book “Calculus Demystified”. However I don’t understand the result of the derivative of [ e^x^2].

    ( ^ means raised to the power )

    In my calculation I did this:

    If (f ◦ g)(x) = e^x^2 and f(s) = s^2

    Considering that ((f ◦ g)(x))’ =[ df/ds(g(x))].d/dx(f(x))

    then ((f ◦ g)(x))’ = 2(e^x) . (e^x) = 2e^2x


    But the book says that the resulting derivative is 2x.e^x^2.

    If the book is correct, could someone explain how this derivative was calculated ?

  2. #2
    Frenzied Member zaza's Avatar
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    Re: Derivative – what is wrong in my calculation ?

    Welcome to the Forums


    If f is a function of s and s is a function of x, then:

    df/dx = df/ds . ds/dx

    Let's say f(s) = es

    and s(x) = x2

    i.e. f(x) = ex2

    Then df/ds = es - that's how the exponential behaves

    ds/dx = 2x

    So df/dx = 2xex2





    What you are doing is considering the derivative of (ex)2, which is equivalent to taking the derivative of e2x, and that as you have correctly pointed out is 2e2x.


    Make sense?



    zaza
    Last edited by zaza; Jul 7th, 2007 at 05:39 PM.
    I use VB 6, VB.Net 2003 and Office 2010



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