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Derivatives, chain rule 6

  1. Nov 10, 2013 #1
    1. The problem statement, all variables and given/known data

    $$xy+e^y=e$$ Find the value of y'' at the point where x=0.

    2. Relevant equations

    Chain rule

    3. The attempt at a solution

    I find y' first.

    $$x\frac{dy}{dx}+y+e^y\frac{dy}{dx}=0$$ $$(x+e^y)\frac{dy}{dx}=-y$$ $$\frac{dy}{dx}=\frac{-y}{x+e^y}$$

    Then I find y''.

    $$\frac{(x+e^y)(-\frac{dy}{dx})-(-y)(1+e^y\frac{dy}{dx})}{(x+e^y)^2}$$ $$\frac{(x+e^y)(\frac{y}{x+e^y})-(-y-ye^y(\frac{-y}{x+e^y}))}{(x+e^y)^2}$$ $$\frac{y-(-y+\frac{y^2e^y}{x+e^y})}{(x+e^y)^2}$$ $$\frac{y+y-\frac{y^2e^y}{x+e^y}}{(x+e^y)^2}$$ $$(2y-\frac{y^2e^y}{x+e^y})(\frac{1}{(x+e^y)^2})$$ $$\frac{2y(x+e^y)-y^2e^y}{(x+e^y)^3}$$ $$\frac{2xy+2ye^y-y^2e^y}{(x+e^y)^3}$$

    x=0 so plugging that in I get

    $$\frac{ye^y(2-y)}{e^{3y}}=\frac{y(2-y)}{e^{2y}}$$

    The answer is $$\frac{1}{e^2}$$ What did I do wrong?

    Any help is much appreciated.
     
  2. jcsd
  3. Nov 10, 2013 #2

    Dick

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    You aren't finished. Same issue as in the last problem. Did you try using your function definition to simplify that?
     
  4. Nov 10, 2013 #3
    I've tried plugging in $$e^y=e-xy$$ but I don't get anywhere.

    $$\frac{y(2-y)}{e^{2y}}=\frac{y(2-y)}{(e-xy)^2}$$

    x=0 so

    $$\frac{y(2-y)}{e^2}$$
     
  5. Nov 10, 2013 #4

    Dick

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    Put x=0 into your original equation. Can you conclude anything about the value of y when x=0?
     
  6. Nov 10, 2013 #5
    What do you mean?

    If I plug x=0 into my original equation I get $$e^y=e-(0)y$$ so $$e^y=e$$ Or I can rearrange the equation to get $$y=\frac{e-e^y}{x}=\frac{e-e^y}{0}=undefined$$
     
  7. Nov 10, 2013 #6

    Dick

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    You should be able to figure out what y is when x=0 from e^y=e. What is it?
     
  8. Nov 10, 2013 #7
    y=1. Okay, thanks!
     
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