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anonymous

  • 5 years ago

how do you derive the derivative of cotx? I know how to do sin, cos, csc, sec, tan

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  1. anonymous
    • 5 years ago
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    d/dx (cotx) = -csc^2(x)

  2. anonymous
    • 5 years ago
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    how do you derive it? I know what it is.

  3. bahrom7893
    • 5 years ago
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    use u sub: u = Sinx du = Cosx

  4. anonymous
    • 5 years ago
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    oh. My mistake. Well, one way would be to use the quotient rule. So cot(x) = cos(x)/sin(x). Taking the derivative of cos(x)/sin(x) using the quotient rule yields: [sin(x)[d/dx(cos(x)] - cos(x)[d/dx(sin(x))]]/sin^2(x)

  5. bahrom7893
    • 5 years ago
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    rewrite Cot = Cos/Sin

  6. bahrom7893
    • 5 years ago
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    please fan us both if we helped!

  7. anonymous
    • 5 years ago
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    What identity does it become? Thats where i got confused in the quotient rule

  8. bahrom7893
    • 5 years ago
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    Okay it becomes: Integral Cot = Integral Cos/Sin Let u = sin du = Cos dx Integral Cot = Integral du/u = Ln|u| + C = Ln|Sin(x)|+C

  9. anonymous
    • 5 years ago
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    its calc 1, I havent done integrals yet

  10. bahrom7893
    • 5 years ago
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    WOOPS LOL sorry i misread the question

  11. bahrom7893
    • 5 years ago
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    Use Cot = Cos*(Sin)^(-1) And use product rule

  12. anonymous
    • 5 years ago
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    from where I left off: the top becomes -sin^2(x) - cos^2(x) = -1 then the whole thing becomes -1/sin^2(x) = -csc^2(x)

  13. bahrom7893
    • 5 years ago
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    Or Cot = Cos/Sin and quotient rule

  14. anonymous
    • 5 years ago
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    beautiful, thanks pyeh9

  15. anonymous
    • 5 years ago
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    my professor wanted us to derive all identities so I was confused on that one for the exam tomorrow

  16. anonymous
    • 5 years ago
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    sure thing!

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