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SandyHearts Group Title

Find the derivative: sin(xcosx) so the answer is after being simplified (cosx-xsinx)cos(xcosx) I have no idea. how.

  • 2 years ago
  • 2 years ago

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  1. Outkast3r09 Group Title
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    this is chain rule

    • 2 years ago
  2. Outkast3r09 Group Title
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    \[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\] where your y=sin(u) and your u=xcos(x)

    • 2 years ago
  3. SandyHearts Group Title
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    ok so: sin(xcosx)(-sinx)(1)+cos(xcosx)

    • 2 years ago
  4. SandyHearts Group Title
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    why is that not the answer I am getting :(

    • 2 years ago
  5. Outkast3r09 Group Title
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    alright so derivative of sin(u) = cos(u) so \[\frac{dy}{du}=cos(u)\] next do the product rule with your u=xcos(x) u'=vw'+wv'=-xsin(x)+cos(x)

    • 2 years ago
  6. Outkast3r09 Group Title
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    so you end up with cos(xcos(x)(-xsin(x)+cos(x))

    • 2 years ago
  7. Outkast3r09 Group Title
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    which is the same as (cos(x)-xsin(x))cos(xcos(x))

    • 2 years ago
  8. Outkast3r09 Group Title
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    does this make sense?

    • 2 years ago
  9. meera_yadav Group Title
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    hi |dw:1340953644460:dw|

    • 2 years ago
  10. SandyHearts Group Title
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    from this cos(xcos(x)(-xsin(x)+cos(x)) all i can get is cos(xcos(x)(-xsin(x)) where does the +cos(x) come from

    • 2 years ago
  11. myininaya Group Title
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    \[\cos( x \cos(x)) \cdot (x \cos(x))' \text{ by chain rule } \]

    • 2 years ago
  12. myininaya Group Title
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    Now use product rule for that one part

    • 2 years ago
  13. SandyHearts Group Title
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    OHHHHHHHH

    • 2 years ago
  14. SandyHearts Group Title
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    >.<

    • 2 years ago
  15. SandyHearts Group Title
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    I see @Outkast3r09 tried explaining that to me but I was really not understanding all the notations you used. Thank you all so much.

    • 2 years ago
  16. SandyHearts Group Title
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    Sorry @Outkast3r09 I didn't understand the first time you explained that to me.

    • 2 years ago
  17. Outkast3r09 Group Title
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    it comes from the chain rule do you understand this \[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\] if you name your inner function u and your outer function y, your outer function will become y=f(u).. so now you find the derivative of your y in respects to u

    • 2 years ago
  18. Outkast3r09 Group Title
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    then you take your u which is u=f(x)... and find the derivative in respects to x.. then multply them together

    • 2 years ago
  19. Outkast3r09 Group Title
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    since your u=xcos(x) you have to use product rule to find the derivative

    • 2 years ago
  20. SandyHearts Group Title
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    I'm going to try to work it out again

    • 2 years ago
  21. SandyHearts Group Title
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    y=sin(xcosx) y'= cos(xcosx)(x)(-sinx)+cosx That's the same thing right? I think I got it.

    • 2 years ago
  22. SandyHearts Group Title
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    idek how i got that anymore.

    • 2 years ago
  23. Outkast3r09 Group Title
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    hehe alright start by turning your outer function into y=f(u)

    • 2 years ago
  24. Outkast3r09 Group Title
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    by letting your u be the innre function

    • 2 years ago
  25. dpaInc Group Title
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    ok... let's start from the beginning.. \[\huge y=sin(xcosx) \] \[\huge y'=cos(xcosx)\cdot [xcosx]' \] by the chain rule...

    • 2 years ago
  26. Outkast3r09 Group Title
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    i'd start by using the actual theorem dpalnc

    • 2 years ago
  27. dpaInc Group Title
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    now... \[\huge [xcosx]'=x'\cdot cosx + x \cdot [cosx]' \] \[\huge [xcosx]'=1\cdot cosx + x \cdot (-sinx) \] \[\huge [xcosx]'= cosx - x sinx \] by the product rul...

    • 2 years ago
  28. dpaInc Group Title
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    now putting these together...

    • 2 years ago
  29. dpaInc Group Title
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    \[\huge y'=cos(xcosx)\cdot (cosx - xsinx) \]

    • 2 years ago
  30. SandyHearts Group Title
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    THANK YOU!!!!!

    • 2 years ago
  31. SandyHearts Group Title
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    I got it. 100%

    • 2 years ago
  32. dpaInc Group Title
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    yw...:)

    • 2 years ago
  33. dpaInc Group Title
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    just take it one step at a time and you should be good...:)

    • 2 years ago
  34. SandyHearts Group Title
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    :) Alright I will.

    • 2 years ago
  35. dpaInc Group Title
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    :)

    • 2 years ago
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