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Homework Help: Derivatives, chain rule 4

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of $$y=cos(\frac{1-e^{2x}}{1+e^{2x}})$$

    2. Relevant equations

    Chain rule

    3. The attempt at a solution

    $$y=cosu$$ $$\frac{dy}{du}=-sinu$$

    $$u=\frac{1-e^{2x}}{1+e^{2x}}$$ $$ \frac{du}{dx}=(1-e^{2x})(-(1+e^{2x})^{-2})+(1+e^{2x})^{-1}(-2e^{2x})$$ $$= -\frac{1-e^{2x}}{(1+e^{2x})^2} + \frac{-2e^{2x}}{1+e^{2x}}$$ $$= \frac{-(1-e^{2x})}{(1+e^{2x})^2} + \frac{(-2e^{2x})(1+e^{2x})}{1+e^(1+e^{2x})^2}$$ $$= -\frac{-1+e^{2x}+(-2e^{2x})+(-2e^{4x})}{(1+e^{2x})^2}$$

    $$\frac{dy}{dx}=-sin\frac{1-e^{2x}}{1+e^{2x}} (-\frac{-1+e^{2x}+(-2e^{2x})+(-2e^{4x})}{(1+e^{2x})^2})$$

    The front part of my answer is right $$-sin\frac{1-e^{2x}}{1+e^{2x}}$$, but the second half is wrong.

    According to my textbook it is supposed to be $$\frac{4e^{2x}}{(1+e^{2x})^2}$$. What did I do wrong?

    Any help is much appreciated.
  2. jcsd
  3. Nov 9, 2013 #2


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    ##=(1-e^{2x})(-(1+e^{2x})^{-2})##<something missing here>##+(1+e^{2x})^{-1}(-2e^{2x})##
    Sign error.
  4. Nov 9, 2013 #3


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    Why don't you just try using the quotient rule instead of the product rule? It's much cleaner, and using the product rule you made a mistake.

    Edit:: Are you trying to wing yourself off using u sub for these yet? It helps, since I believe you forgot [itex] (1+e^{2x})^{-1} [/itex] is also a composition.
    Last edited: Nov 9, 2013
  5. Nov 10, 2013 #4


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    "wing yourself off"? Oh, you mean "wean". That stopped me for a moment.
  6. Nov 10, 2013 #5


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    There are several ways to approach these problems

    I would be inclined to rewrite the function
    $$\cos \left( \frac{1-e^{2x}}{1+e^{2x}} \right)=\cos \left(1- \frac{2}{1+e^{2x}} \right)=\cos(\tanh(x))$$

    Otherwise use the chain rule multiple times with care. For example

    $$\cos \left( \frac{1-e^{2x}}{1+e^{2x}} \right)=\mathrm{f}(\mathrm{g}(\mathrm{h}(x))) \\
    \text{where} \\
    \mathrm{f}(x)=\cos(x) \\
    \mathrm{g}(x)=\frac{1-x}{1+x} \\
    \mathrm{h}(x)=e^{2x} $$
  7. Nov 10, 2013 #6
    Thanks everyone! I've got it!
  8. Nov 10, 2013 #7


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