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calyne Group Title

Find dy/dx by implicit differentiation: e^y cos(x) = 1 + sin(xy)

  • 2 years ago
  • 2 years ago

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  1. calyne Group Title
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    i got [sin(y) cos(x) + e^y sin(x)] / [e^y cos(x) - sin(x) cos (y)] but the textbook answer is [e^y sin(x) + y cos(xy)] / [e^y cos(x) - x cos(xy)] so am i close or am i there and if those answers are the same how are they equal show me

    • 2 years ago
  2. eseidl Group Title
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    \[y'e^y \cos x -e^y \sin x=0+\cos(xy)(x)y'+\cos(xy)y\]so,\[y'=\frac{e^ysinx+\cos(xy)y}{e^ycosx-xcos(xy)}\]

    • 2 years ago
  3. eseidl Group Title
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    not sure where you are getting siny*cosx terms...they are incorrect

    • 2 years ago
  4. calyne Group Title
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    why is d/dx [sin(xy)] equal to cos(xy)(x)y' + cos(xy)y ???? and i know the derivative of cos x, i had it as negative

    • 2 years ago
  5. calyne Group Title
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    i thought it was like sin(x)*sin(y)....... so that's how i tried to get the derivative oops but i still don't get what it is then

    • 2 years ago
  6. eseidl Group Title
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    using the product rule I will differentiate d/dx [sin(xy)] = d/dx [sin(xy)]*d/dx[xy]

    • 2 years ago
  7. eseidl Group Title
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    we get: cos (xy)*y

    • 2 years ago
  8. calyne Group Title
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    no ok i know how

    • 2 years ago
  9. calyne Group Title
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    but sin isn't its own fluttering thing

    • 2 years ago
  10. calyne Group Title
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    oh is it

    • 2 years ago
  11. calyne Group Title
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    oh wait flutter i'm reallll rusty on my trig i never learned it properly professor never got around to it so i had to cram myself for the departmental exam

    • 2 years ago
  12. calyne Group Title
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    so yeah thanks anyway

    • 2 years ago
  13. calyne Group Title
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    idk

    • 2 years ago
  14. eseidl Group Title
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    this is the first term. but we have to do it again with respect to y and apply to chain rule so, the second term is:\[d/dx[\sin(xy)]=\cos(xy)*d/dy(xy)*dy/dx\]

    • 2 years ago
  15. eseidl Group Title
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    looks messy but, you get:\[xcos(xy)*y'\]

    • 2 years ago
  16. eseidl Group Title
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    now you sum them (this is just the usual product rule).

    • 2 years ago
  17. eseidl Group Title
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    so really we are applying the product rule AND the chain rule to get this

    • 2 years ago
  18. calyne Group Title
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    wait WHAT the flutter d/dx (xy) is x*d/dx(y)(* dy/dx)+y*d/dx(x) so it's x(1) dy/dx + y(1) soooo the answer would be xcos(xy) dy/dx + ycos(xy)

    • 2 years ago
  19. eseidl Group Title
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    as an easier example of this process consider differentiating xy implicitly. d/dx[xy]=y+(d/dy)(dy/dx)(xy) =y+xy'

    • 2 years ago
  20. calyne Group Title
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    why

    • 2 years ago
  21. eseidl Group Title
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    maybe it's easier if we write xy=xf(x) where y= f(x) now differentiate it: d/dx[x*f(x)]=f(x)+xf'(x) or, =y+xy'

    • 2 years ago
  22. eseidl Group Title
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    this is just the product rule, right? when we implicitly differentiate we assume that y is a function of x. y=f(x)

    • 2 years ago
  23. calyne Group Title
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    yeah yeah but flutter I DON"T EVEN KNOW THE flutterING IMPLICIT DIFFERENTIATION RULES MY TEXTBOOK SUCKS AND MY PROFESSOR SUCKS AND THERE ARE NO flutterING RULES IN THE TEXTBOOK FOR SOME REASON IT JUST SHOWS EXAMPLES WITHOUT EVEN EXPLAINING so ALL I KNOW is that you stick y' to multipy the derivative of any term with a y in it

    • 2 years ago
  24. calyne Group Title
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    that's fluttering it i don't know why or how or the details or fluttering anything flutter

    • 2 years ago
  25. eseidl Group Title
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    when we have x*f(x) the overall change must take into account the rate of change of x and the rate of change of f(x). the product rule says that the overall rate of change of x*f(x) is the rate of change of x (i.e., =1) time f(x) PLUS x*(the rate of change of f(x)=f'(x))

    • 2 years ago
  26. calyne Group Title
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    and wtf so the product rule for sin(xy) how does that work too

    • 2 years ago
  27. calyne Group Title
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    product rule is d/dx[uv] = u d/dx[v] + v d/dx (u). wtf is u and what is v first of all.

    • 2 years ago
  28. eseidl Group Title
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    remember we work from the outside to the inside. if we has sin(8x), working outside in, we get cos(8x)*d/dx(8x)=8cosx.

    • 2 years ago
  29. calyne Group Title
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    in sin(xy) how does that translate

    • 2 years ago
  30. calyne Group Title
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    but xy is x*y

    • 2 years ago
  31. calyne Group Title
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    ugh

    • 2 years ago
  32. eseidl Group Title
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    yeah, the only difference between sin(8x) and sin(xy) is y is a function of x and 8 is not

    • 2 years ago
  33. eseidl Group Title
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    so we have to account for this in the overall rate of change. sin(xy)=cos(xy)*(d/dx)(xy)

    • 2 years ago
  34. eseidl Group Title
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    now...the crucial thing to realize here is both x and y are functions of x in the above

    • 2 years ago
  35. eseidl Group Title
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    so we have cos(xy) times d/dx(xy). we use the product rule on this second term. d/dx(xy)=y+xy' So overall we get cos(xy)[y+xy']=ycos(xy)+xy'cos(xy

    • 2 years ago
  36. eseidl Group Title
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    got to go, hope this helps!

    • 2 years ago
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