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anonymous

  • 5 years ago

derivative!! help!!::: y=2sinx-5cosx

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  1. anonymous
    • 5 years ago
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    Do one term at a time.

  2. anonymous
    • 5 years ago
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    i know for stuff like :x^3+x^2, it's just = 3x^2+2x but how would you do it one by one for SIN and COS??

  3. anonymous
    • 5 years ago
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    derivative of sin= cos derivative of cos=-sin

  4. anonymous
    • 5 years ago
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    Is the first part -2cosx?

  5. anonymous
    • 5 years ago
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    ohh, so it's just -2cosx +5sin?

  6. anonymous
    • 5 years ago
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    You have the derivatives in my note above. Please note the signs.

  7. anonymous
    • 5 years ago
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    aaak I subtracted should be added: f'g + g'f f=2 f'=0 g=sinx g'=cos x 0*sinx + cosx * 2 but for reasons known only to my brain I wanted to subtract so first part is 2 cosx no negativity

  8. anonymous
    • 5 years ago
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    bad kantalope! trying to do things in head

  9. anonymous
    • 5 years ago
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    No multiplication rule.

  10. anonymous
    • 5 years ago
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    no? uh oh....what rule should I be usin?

  11. anonymous
    • 5 years ago
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    Since they are added, you just find the derivative of each.

  12. anonymous
    • 5 years ago
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    right so 2sinx is two times sine x ya? how come no multiplication rule? Ah because it is not (2x)sinx......so should treat it like coefficient in 3x^2

  13. anonymous
    • 5 years ago
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    Right 2 is just a coefficient, it goes along for the ride, as they say.

  14. anonymous
    • 5 years ago
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    well dang it - that explains some of my test scores lol

  15. anonymous
    • 5 years ago
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    It takes time to sink in. You've got it now.

  16. anonymous
    • 5 years ago
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    this is the answer 2cosx+5sinx

  17. anonymous
    • 5 years ago
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    i hope my memory did not tell me wrong.:)lol

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