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anonymous

  • 5 years ago

Find dy/dx by implicit differentiation. 5+3x=sin(xy)^2

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  1. anonymous
    • 5 years ago
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    d/dx(5+3x) = d/dx(sin(xy)^2 so the left side is pretty obvious...I'm guessing its the RHS that's giving you trouble.

  2. anonymous
    • 5 years ago
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    \[d/dx \sin (xy)^2 = d/dx sinu *d/dx (u) \] if we use the chain rule and (yx)^2=u

  3. anonymous
    • 5 years ago
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    i got cos(xy)^2*2xy* 1*y*y' for that part but idk if its right

  4. anonymous
    • 5 years ago
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    so d/dx (u) is where implicit differentiation comes in. You'll have to use two rules: the product rule of differentiation (d/dt (xy) = (x)'*y + x*(y)' ) and the implicit differentiation rule.

  5. anonymous
    • 5 years ago
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    i no how to do the product rule, its just the implicit differientiation rule that confuses me

  6. amistre64
    • 5 years ago
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    rules for implicit are exactly the same for explicit; only thing is you want to keep your x' and y' til the end then factor out your y'. your x'[dx/dx] will of course equal 1

  7. anonymous
    • 5 years ago
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    so when u differenciate the y wat does that become? y*y'?

  8. amistre64
    • 5 years ago
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    yes, y becomes y' 6y^2 becomes 12y y'

  9. anonymous
    • 5 years ago
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    oooooooooooooooooooo!!!!

  10. amistre64
    • 5 years ago
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    xy becomes: x'y + xy' not that it helps here....maybe

  11. amistre64
    • 5 years ago
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    ...but it might lol

  12. anonymous
    • 5 years ago
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    hmmm...I was thinking it would, since d(xy)/dx) = y+x*dy/dx?

  13. amistre64
    • 5 years ago
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    2(sin(xy)) cos(xy) (x'y + xy') ??

  14. anonymous
    • 5 years ago
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    so when i differenciate the whole thing it becomes 0+3=cos(xy)^2 * 2xy* (1)(y)(y')?

  15. anonymous
    • 5 years ago
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    or not...

  16. amistre64
    • 5 years ago
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    u^2 Du ; u = sin(t); t = xy

  17. amistre64
    • 5 years ago
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    2sin(xy) cos(xy) (x'y + xy') does that help out?

  18. anonymous
    • 5 years ago
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    yaa so thats for the right side only right

  19. anonymous
    • 5 years ago
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    so its a chain rule within a chain rule?

  20. amistre64
    • 5 years ago
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    yep; chain a chain lol

  21. amistre64
    • 5 years ago
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    D(u) D(sin(t)) D(t) is what I get from it

  22. anonymous
    • 5 years ago
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    ooooooo ok i get it!

  23. anonymous
    • 5 years ago
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    so now i solve for y'

  24. amistre64
    • 5 years ago
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    yep, x' = 1 so we can ignore those; and solve for y'

  25. anonymous
    • 5 years ago
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    do i divide 2sinxy*cos(xy) on both sides?

  26. amistre64
    • 5 years ago
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    its a good start :)

  27. amistre64
    • 5 years ago
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    then subtract a "y" and divide it all again by "x"

  28. anonymous
    • 5 years ago
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    so i got 3-y/2sin(xy)*cos (xy)*x

  29. anonymous
    • 5 years ago
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    is that correct?

  30. amistre64
    • 5 years ago
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    thats a little rough, not quite it....

  31. anonymous
    • 5 years ago
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    awwww!!! lol

  32. anonymous
    • 5 years ago
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    wat did i do wrong?

  33. amistre64
    • 5 years ago
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    3 y ---------------- - -- x (2sin(xy)cos(xy)) x

  34. anonymous
    • 5 years ago
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    where did the x under the 3 come from?

  35. amistre64
    • 5 years ago
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    lets say B = (2sin(xy)cos(xy)) to clean this up... 3/B = y + xy' (3/B) -y = xy' (3/B)/x - y/x = y' 3/Bx - y/x = y'

  36. anonymous
    • 5 years ago
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    oo i turn the x into a 1

  37. amistre64
    • 5 years ago
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    x' = dx/dx = 1 like any good fracation does :)

  38. anonymous
    • 5 years ago
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    ooooooooo ok i see wat i did wrong

  39. amistre64
    • 5 years ago
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    you got 3-y/Bx

  40. anonymous
    • 5 years ago
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    omg im gonna do exactly wat u did for the test.. change that whole part to a B so i wont make a mistake

  41. amistre64
    • 5 years ago
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    lol ..... it is easier on the eyes fer sure :)

  42. anonymous
    • 5 years ago
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    THANK U SOO MUCH!!!

  43. anonymous
    • 5 years ago
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    UR A LIFE SAVER!!

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