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SandyHearts

  • 2 years ago

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

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  1. Outkast3r09
    • 2 years ago
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    this is chain rule

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

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

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

  5. Outkast3r09
    • 2 years ago
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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)

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

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

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

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

  10. SandyHearts
    • 2 years ago
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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

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

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

  13. SandyHearts
    • 2 years ago
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    OHHHHHHHH

  14. SandyHearts
    • 2 years ago
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    >.<

  15. SandyHearts
    • 2 years ago
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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.

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

  17. Outkast3r09
    • 2 years ago
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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

  18. Outkast3r09
    • 2 years ago
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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

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

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

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

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

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

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

  25. dpaInc
    • 2 years ago
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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...

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

  27. dpaInc
    • 2 years ago
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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...

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

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

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

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

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

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

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

  35. dpaInc
    • 2 years ago
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    :)

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