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Homework Help: Finding y' by implicit differentiation SO HARD FOR ME! help~

  1. Jan 3, 2005 #1
    Find dy/dx by implicit differentiation. (Meaning find the derivative of y.. so y' = ??)

    cos(x-y)=xe^x

    Please help and show step by step..
    the final answer should be..

    y' = 1 + [(e^x)(1+x)]/[sin(x+y)]

    Thanks a bunch. :smile:
     
  2. jcsd
  3. Jan 3, 2005 #2

    dextercioby

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    If you know that the LHS is a function of "x" and the RHS is a function of "x" as well,then the equality A=B implies A'=B' (the equality of derivatives wrt to "x").So differentiate both sides of the eq.wrt to "x":
    [tex] -[\sin(x-y)](1-\frac{dy}{dx})=e^{x}+xe^{x}\Rightarrow\frac{dy}{dx}-1=\frac{e^{x}(x+1)}{\sin(x-y)}\Rightarrow\frac{dy}{dx}=\frac{e^{x}(x+1)}{\sin(x-y)}+1 [/tex]
    ,which means that your answer was wrong.

    Daniel.
     
  4. Jan 3, 2005 #3

    HallsofIvy

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    It's just the chain rule. Since y is a function of x, the derivative of cos(x-y) is
    -sin(x-y) times the derivative of x-y, with respect to x. Of course, the derivative of x is 1. We don't know y as a function of x so we can only write its derivative as y':
    (cos(x-y))'= -sin(x-y)(1- y').

    The derivative of xex is (by the product rule) ex+ xex = ex(1+ x) so we have

    -sin(x-y)(1- y')= ex(1+ x).

    Now solve for y'.
     
  5. Jan 3, 2005 #4
    Oh, Thanks. I did it so that , cos(x-y) = cosx/cosy, so i did : (-cosysinx + cosxsinyy')/(cos^2 y). Is that a valid method of doing that?
     
  6. Jan 4, 2005 #5

    HallsofIvy

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    Who were you responding to? Neither dexercioby nor I had "cos(x-y)" in our formulas.

    And where did you get "cos(x-y)= cos(x)/cos(y)" from?
     
  7. Jan 4, 2005 #6
    I meant when i tried solving it, i put it so that cos(x-y) is simplified to cosx/cosy, then i tried deriving that then solve the rest. I guess i did it wrong. Sorry for the confusion.
     
  8. Jan 4, 2005 #7

    dextercioby

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    I hope for your sake that u are convinced that
    [tex] \cos(x-y)\neq \frac{\cos x}{\cos y} [/tex]
    ,but instead
    [tex] \cos(x-y)=\cos x\cos y+\sin x\sin y [/tex]

    Daniel.
     
  9. Jan 4, 2005 #8
    Oh thanks. I just looked up in the book and thats it! Thanks.
     
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