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Math Help - another derivative

  1. #1
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    another derivative

    How would you do
    y = sin^-2 (x)/(1+x^2)
    find yprime
    I got an answer of (1+x^2)(-2sin^-1 (x) - 2xsin^-2 (x)/(1+x^2)^2
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  2. #2
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    You say \sin^{-2}x=\arcsin^2x isn't it?
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    You say \sin^{-2}x=\arcsin^2x isn't it?
    Yeah i think those are equal.
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  4. #4
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    Hello, davecs77!

    Differentiate: . y \:=\:\frac{(\sin x)^{-2}}{1+x^2}

    I would re-write it as: . y \;=\;\frac{\csc^2\!x}{1+x^2}

    Then: . y' \;=\;\frac{(1+x^2)\cdot2\csc x\cdot(-\csc x\cot x) - \csc^2\!x\cdot(2x)}{(1+x^2)^2}

    . . . . . y' \;=\;\frac{-2(1+x^2)\csc^2\!x\cot x - 2x\csc^2\!x}{(1+x^2)^2}

    . . . . . y' \;=\;\frac{-2\csc^2\!x\cdot\left[(1+x^2)\cot x - x\right]}{(1+x^2)^2}

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