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Math Help - Differentiation:very confused...please help!

  1. #1
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    Differentiation:very confused...please help!

    Hi, I have this diff. question which i keep getting wrong:

    Find dy/dx of:

    v=1/(1+e^-x)

    This is my working:

    v=1/1 + 1/(e^-x)=1+1/(e^x)=1+e^x

    therefore dy/dx would equal :

    e^x

    could someone please show me where i'm going wrong...

    Thank you
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Kris View Post
    Hi, I have this diff. question which i keep getting wrong:

    Find dy/dx of:

    v=1/(1+e^-x)

    This is my working:

    v=1/1 + 1/(e^-x)=1+1/(e^x)=1+e^x

    therefore dy/dx would equal :

    e^x

    could someone please show me where i'm going wrong...

    Thank you
    I'm curious as to how you got to calculus 1 without knowing how to deal with fractions.


    Firstly,
    In general: a/(b + c) IS NOT EQUAL TO a/b + a/c

    so 1/(1+e^-x) IS NOT THE SAME AS 1/1 + 1/e^-x

    we can only split up a fraction in that way if we have a single term in the denominator.
    for example, (a + b)/c = a/c + b/c

    so first thing you need to do is review fractions.

    Secondly,
    dy/dx means "the derivative of y with respect to x" there is no y here. you wan dv/dx, "the derivative of v with respect to x"


    Now to your problem


    v = 1/(1+e^-x) ............change this in terms of an exponent
    => v = (1 + e^-x)^-1 ......now we proceed by the Chain rule

    Note: Chain Rule
    [f(g(x))]' = f'(g(x))*g'(x)

    So, v = (1 + e^-x)^-1
    => dv/dx = -1(1 + e^-x)^-2 *(1 + e^-x)'
    .........= -(1 + e^-x)^-2 * (-e^-x)
    .........= (e^-x)(1 + e^-x)^-2
    .........= (e^-x)/[(1 + e^-x)^2]
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  3. #3
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    thank you, that was just a stupid mistake about the fractions, wasn't thinking straight yesterday and just couldn't figure it!
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