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anonymous

  • one year ago

Find an equation of the tangent line to the curve at the given point. y = sin 5x + sin^2 5x, (0, 0) I got 5x but it was wrong please help me

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  1. Loser66
    • one year ago
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    y'= ??

  2. anonymous
    • one year ago
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    cos5x+2sin5x?

  3. anonymous
    • one year ago
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    Forgot chain rule.

  4. Loser66
    • one year ago
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    it's not that but I got what you have. hahahaha....

  5. anonymous
    • one year ago
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    \(y' = cos(5x)\cdot5 +2sin(5x)\cdot cos(5x)\cdot5\) \(y' = 5cos(5x) + 10sin(5x)cos(5x)\)

  6. Loser66
    • one year ago
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    @Concentrationalizing some more steps, you will get what he got and it is wrong (according to him) :)

  7. Loser66
    • one year ago
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    but I don't know how it is wrong!! or you should write y = 5x @gaba

  8. anonymous
    • one year ago
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    Plug in 0 into the derivative: \(y' = 5cos(0) + 10sin(0)cos(0)\) \(y' = 5\) Hmm....question is typed incorrectly?

  9. anonymous
    • one year ago
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    i needed to type it as y=5x! thank you!!

  10. anonymous
    • one year ago
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    oh, lol. Okay, cool.

  11. Loser66
    • one year ago
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    hahahaha...................... it's fun.

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spraguer (Moderator)
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