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Math Help - derivative

  1. #1
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    derivative

    I am trying to take the derivative of this:
    y = arccsc^2(x)
    I know this can be simplied as arccsc(x)^2
    I am trying to find y(prime)..can I use natural logs to do that?
    Could I do lny = 2ln(arccsc(x))
    Thanks for the feedback.
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  2. #2
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    You could perhaps try it this way:

    \frac{d}{dx}[csc^{-1}(x)]

    y=csc^{-1}(x), \;\ x=csc(y)

    Differentiate:

    \frac{d}{dx}{x}=\frac{d}{dx}[csc(y)]\Rightarrow{1=-csc(y)cot(y)}\frac{dy}{dx}

    \frac{dy}{dx}=\frac{1}{-csc(y)cot(y)}

    \frac{dy}{dx}=\frac{1}{-csc(csc^{-1}(x))cot(csc^{-1}(x))}

    \frac{dy}{dx}=\frac{-1}{x\sqrt{x^{2}-1}}


    \frac{d}{dx}[csc^{-1}(x)^{2}]

    Chain rule:

    2csc^{-1}(x)\frac{d}{dx}[csc^{-1}(x)]

    \boxed{\frac{-2csc^{-1}(x)}{x\sqrt{x^{2}-1}}}
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  3. #3
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    Quote Originally Posted by galactus View Post
    You could perhaps try it this way:

    \frac{d}{dx}[csc^{-1}(x)]

    y=csc^{-1}(x), \;\ x=csc(y)

    Differentiate:

    \frac{d}{dx}{x}=\frac{d}{dx}[csc(y)]\Rightarrow{1=-csc(y)cot(y)}\frac{dy}{dx}

    \frac{dy}{dx}=\frac{1}{-csc(y)cot(y)}

    \frac{dy}{dx}=\frac{1}{-csc(csc^{-1}(x))cot(csc^{-1}(x))}

    \frac{dy}{dx}=\frac{-1}{x\sqrt{x^{2}-1}}


    \frac{d}{dx}[csc^{-1}(x)^{2}]

    Chain rule:

    2csc^{-1}(x)\frac{d}{dx}[csc^{-1}(x)]

    \boxed{\frac{-2csc^{-1}(x)}{x\sqrt{x^{2}-1}}}
    I'm not really understanding how do did this. You found the derivative of arccsc(x), (1/-x((squareroot(x^2 - 1)) But how did you figure out yprime from y = arccsc^2(x)?
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  4. #4
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    The last part. The chain rule.
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  5. #5
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    Ohh..ok
    Where i am getting lost is how are you breaking y = arccsc^2 (x) up?
    I see you take the derivative of arccsc(x) (which you set to y)
    but where is the (csc^-1 (x)^2) coming from?
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