Results 1 to 5 of 5

Math Help - differentiation of e using chain rule

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    74

    differentiation of e using chain rule

    Could someone assist with this, please? The task is to differentiate the following equation

    y=xe^{-kx}

    by the chain rule, differentiation should be e^{f(x)}*f'(x) for e, so to me that means

    y'=xe^{-kx}(-k) or y'=-kxe^{-kx}

    unfortunately, my book would disagree with me. According to it, the answer before simplification is

    y'=xe^{-kx}(-k) + e^{-kx}*1

    Can someone please explain to me where the plus and everything after it came from?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    You need to use the product rule on the expression as a whole

    If y = u(x)v(x) then y' = u'v + v'u

    You should set u=x and v = e^{-kx}

    Use the chain rule to find v'
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by satis View Post
    Could someone assist with this, please? The task is to differentiate the following equation

    y=xe^{-kx}

    by the chain rule, differentiation should be e^{f(x)}*f'(x) for e, so to me that means

    y'=xe^{-kx}(-k) or y'=-kxe^{-kx}

    unfortunately, my book would disagree with me. According to it, the answer before simplification is

    y'=xe^{-kx}(-k) + e^{-kx}*1

    Can someone please explain to me where the plus and everything after it came from?
    1. The given function is a product of functions. That means you have to use the product rule:

    f(x)=u(x) \cdot v(x)~\implies~ f'(x)=v(x) \cdot u'(x) + u(x) \cdot v'(x)

    2. With your function you have u(x) = x, v(x) = e^{-kx}

    3. The first step should be:

    f'(x)=e^{-kx} \cdot 1 + x \cdot \underbrace{e^{-kx} \cdot (-k)}_{chain-rule}

    4. Simplify!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2009
    Posts
    74
    ahhh, got it. I'd totally glossed over the fact I had a product of two functions. thanks to both of you for pointing out the problem.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by satis View Post
    Could someone assist with this, please? The task is to differentiate the following equation

    y=xe^{-kx}

    by the chain rule, differentiation should be e^{f(x)}*f'(x) for e, so to me that means

    y'=xe^{-kx}(-k) or y'=-kxe^{-kx}

    unfortunately, my book would disagree with me. According to it, the answer before simplification is

    y'=xe^{-kx}(-k) + e^{-kx}*1

    Can someone please explain to me where the plus and everything after it came from?


     y' = (x)' e^{-kx} + x (e^{-kx} )'

    Edit: sorry ... to late
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differentiation and Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 7th 2010, 05:05 AM
  2. Chain Rule Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 26th 2010, 02:03 PM
  3. chain rule + differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 25th 2010, 02:16 PM
  4. Replies: 2
    Last Post: September 5th 2009, 11:26 PM
  5. differentiation - chain rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 9th 2008, 07:14 AM

/mathhelpforum @mathhelpforum