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anonymous

  • 5 years ago

How do I find the second derivative of -1/sin ^2x ?

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  1. anonymous
    • 5 years ago
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    is it -1/sin^2 (x)

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    i mean sin square x right?

  4. anonymous
    • 5 years ago
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    write 1 as sin^2x+cos^2x

  5. anonymous
    • 5 years ago
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    it will become-(1+cot^2x)

  6. anonymous
    • 5 years ago
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    u need help?

  7. anonymous
    • 5 years ago
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    actually i was not able to post it on grp chat

  8. anonymous
    • 5 years ago
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    @zain

  9. anonymous
    • 5 years ago
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    I'd just change it to -csc^2(x) and then take the first derivative, and then chain rule for the second derivative

  10. anonymous
    • 5 years ago
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    or else use uv rule

  11. anonymous
    • 5 years ago
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    Thanks guys

  12. anonymous
    • 5 years ago
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    taking 1=u & sinx =v & integrate

  13. anonymous
    • 5 years ago
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    im sry...differentiate , (not integrate)

  14. anonymous
    • 5 years ago
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    Differentiate by parts!! lol :)

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