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Determine dy/dx of the following and simplify if possible

  1. Aug 12, 2015 #1
    < Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

    cos(x-y)=ysinx

    My attempt:
    -sin(x-y(x'-y')=y'cosx.x'

    Yeah I'm stuck.. I know it is differentiation of implicit functions and I need to make y' the subject of the formula.
     
    Last edited by a moderator: Aug 12, 2015
  2. jcsd
  3. Aug 12, 2015 #2

    ShayanJ

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    Do you know that ## x'=\frac{dx}{dx}=1 ##?
     
  4. Aug 12, 2015 #3
    I do now :smile:
     
  5. Aug 12, 2015 #4

    ShayanJ

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    Actually you've done something wrong. You should have ## (y'-1)\sin(x-y)=y'\sin x+y \cos x \Rightarrow [\sin(x-y)-\sin x]y'=y\cos x +\sin(x-y) ##.
    Can you continue?
     
  6. Aug 12, 2015 #5
    y'= ##\frac{ycosx+sin(x-y)}{sin(x-y)-sinx}##

    y'=##\frac{ycosx}{-sinx}##

    Therefore ##\frac{dy}{dx}## = ##\frac{ycosx}{-sinx}##
     
  7. Aug 12, 2015 #6

    ShayanJ

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    What happened to ##\sin(x-y)##? You can't do that!
     
  8. Aug 12, 2015 #7
    Um it got cancelled because it was the numerator and denominator of the fraction? My bad
     
  9. Aug 12, 2015 #8

    ShayanJ

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    If it was something like ##\frac{f(x)g(x)}{f(x)h(x)}##, then you could cancel f(x) and have ## \frac{g(x)}{h(x)}##. But here you have ##\frac{f(x)+g(x)}{f(x)+h(x)}##. you can't simplify further.
     
  10. Aug 12, 2015 #9
    ##\frac{dy}{dx}## = ##\frac{ycosx+sin(x-y)}{sin(x-y)-sinx}##

    Is the final answer then?
     
  11. Aug 12, 2015 #10

    ShayanJ

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  12. Aug 12, 2015 #11
    Thank you very much :smile:
     
  13. Aug 16, 2015 #12
    Final answer attached. Thanks to all who helped.
     

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