Determine dy/dx of the following and simplify if possible

  • Thread starter DevonZA
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  • #1
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< 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.
 
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Answers and Replies

  • #2
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Do you know that ## x'=\frac{dx}{dx}=1 ##?
 
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  • #3
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Do you know that ## x'=\frac{dx}{dx}=1 ##?
I do now :smile:
 
  • #4
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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?
 
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  • #5
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y'= ##\frac{ycosx+sin(x-y)}{sin(x-y)-sinx}##

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

Therefore ##\frac{dy}{dx}## = ##\frac{ycosx}{-sinx}##
 
  • #6
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What happened to ##\sin(x-y)##? You can't do that!
 
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  • #7
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What happened to ##\sin(x-y)##? You can't do that!
Um it got cancelled because it was the numerator and denominator of the fraction? My bad
 
  • #8
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Um it got cancelled because it was the numerator and denominator of the fraction? My bad
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.
 
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  • #9
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##\frac{dy}{dx}## = ##\frac{ycosx+sin(x-y)}{sin(x-y)-sinx}##

Is the final answer then?
 
  • #11
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Thank you very much :smile:
 
  • #12
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Final answer attached. Thanks to all who helped.
 

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