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Chain Rule

  1. Feb 15, 2017 #1
    1. The problem statement, all variables and given/known data

    I need help finding the derivative. The question wants me to find the equation of the tangent line to the curve

    [itex]y=\dfrac{6}{1+e^{-x}}[/itex] at point [itex](0, 3)[/itex]. I'm unclear on when to use the chain rule at certain areas.

    2. Relevant equations

    Product Rule: [itex]f(x)g'(x)+f'(x)g(x)[/itex]

    Quotient Rule: [itex]\dfrac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}[/itex]

    Chain Rule:

    if [itex]y=g(f(x))[/itex]

    then [itex]\dfrac{dy}{dx}=g'(f(x))\cdot f'(x)[/itex]


    if [itex]y=g(u)[/itex] and [itex]u=f(x)[/itex]

    then [itex]\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}[/itex]

    3. The attempt at a solution




    [itex]y'=\dfrac{(1+e^{-x})(0)-(1+e^{-x})(-e^{-x})(6)}{(1+e^{-x})^2}[/itex] Used chain rule here for [itex]-(1+e^{-x})[/itex]




    Book Answer:









    Everything goes terribly wrong for me when I use the chain rule. Also, at then end, why didn't they square the denominator?
  2. jcsd
  3. Feb 15, 2017 #2


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    There is no chain rule here: just a sum rule if you want: $$(1+e^{-x})' = 1' + (e^{-x})' = 0 - e^{-x} = - e^{-x} $$
  4. Feb 15, 2017 #3
    Thanks, I didn't see that little detail. That helped.
  5. Feb 15, 2017 #4


    Staff: Mentor

    I would never use the quotient rule for a function where the numerator is a constant. I would write ##\frac 6 {1 + e^{-x}}## as ##6(1 + e^{-x})^{-1}## and use the constant multiple rule followed by the chain rule.

    Having said that, your work in the line above is incorrect. You should NOT use the chain rule on ##1 + e^{-x}##, at least not at first. The derivative of ##1 + e^{-x}## is simply ##-e^{-x}##.

    ##\frac d {dx} \left( 1 + e^{-x} \right) = \frac d {dx} 1 + \frac d {dx} e^{-x} = 0 - e^{-x}##
    I used the sum rule followed by the chain rule.
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