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amyna

  • one year ago

find the derivative: y=(x^2+csc(-5x))^3/2 thank you!

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  1. amyna
    • one year ago
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    |dw:1437845982011:dw|

  2. IrishBoy123
    • one year ago
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    \(y=(x^2+cosec(-5x))^{3/2}\) that it? suggest you share what you have done/ what the problem is

  3. anonymous
    • one year ago
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    \[2x^{1}-2/15\csc(-5x)^{-1/3}\]

  4. amyna
    • one year ago
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    i used chain rule and got 3/2(x^2+csc(-5x))^1/2 (2x-csc(-5x)cot(-5x)*5 ? i don't know if thats right?

  5. Loser66
    • one year ago
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    right but + instead of - at the last term. |dw:1437846650378:dw|

  6. IrishBoy123
    • one year ago
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    is this what you are saying? \((3/2)(x^2+csc(-5x))^{1/2} (2x-csc(-5x)cot(-5x).5)\)

  7. amyna
    • one year ago
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    yes

  8. IrishBoy123
    • one year ago
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    look at @Loser66 's sketch \( (cosec \ x)^\prime = - cosec \ x \ cot \ x\)

  9. IrishBoy123
    • one year ago
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    sorry, @loser66 was trying to make this point: \( d/dx (-5x) = -5\). so you have: \( \frac{3}{2}(x^2+cosec(−5x))^{1/2} \ \ (2x−cosec(−5x)cot(−5x)(-5)) \) \(\ = \frac{3}{2}(x^2+cosec(−5x))^{1/2} \ \ (2x+5 \ cosec(−5x)cot(−5x)) \) next, i'd be really tempted to clean out those \(-5x\)'s within the trig functions but maybe you are happy

  10. amyna
    • one year ago
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    thanks! it makes more sense now!

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