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Math Help - Another derivatives of inverse trig function

  1. #1
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    Another derivatives of inverse trig function

    I need help taking the derivative of y = arcsec(2x^2)
    This is what I do
    y = arcsec(2x^2)
    x=sec(2x^2)
    y'=-4ysin(2y^2)

    I am suppose to get a polynomial but I cannot take the derivative of it to get one. I just get stuck at the last step. Any help? This seems not to make any sense at all for me
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  2. #2
    Super Member Quacky's Avatar
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    Re: Another derivatives of inverse trig function

    y = arcsec(2x^2)

    sec(y)=sec(arcsec(2x^2))

    sec(y)=2x^2
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  3. #3
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    Re: Another derivatives of inverse trig function

    Quote Originally Posted by Quacky View Post
    y = arcsec(2x^2)

    sec(y)=sec(arcsec(2x^2))

    sec(y)=2x^2
    with that I get

    y'= \frac{4x}{2x^2tany}

    y'= \frac{2 cot(y)}{x}

    From here I do not know where to go.
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  4. #4
    Super Member Quacky's Avatar
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    Re: Another derivatives of inverse trig function

    Quote Originally Posted by Quacky View Post
    y = arcsec(2x^2)

    sec(y)=sec(arcsec(2x^2))

    sec(y)=2x^2
    Quote Originally Posted by Barthayn View Post
    with that I get

    y'= \frac{4x}{2x^2tany}

    y'= \frac{2 cot(y)}{x}

    From here I do not know where to go.
    Let's see.

    sec(y)=2x^2

    sec(y)tan(y)y'=4x

    y'=\frac{4x}{sec(y)tan(y)}

    y'= \frac{4x}{2x^2tan(y)} as you had (sorry, I had to get there for myself)

    Using the identity:

    tan^2(y)+1=sec^2(y), we get that:

    tan(y)=\sqrt{sec^2(y)-1}

    We had:

    y'= \frac{4x}{2x^2tan(y)}

    y'= \frac{4x}{2x^2\sqrt{sec^2(y)-1}}

    y'= \frac{4x}{2x^2\sqrt{4x^4-1}}

    y'= \frac{2}{x\sqrt{4x^4-1}}
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  5. #5
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    Re: Another derivatives of inverse trig function

    Thanks, I had you third last step and see sec^2y and I was all like "what do I do? Help someone!". Thanks for pointing out the unit trig identity. It was much easier than it appeared
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  6. #6
    Super Member Quacky's Avatar
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    Re: Another derivatives of inverse trig function

    Quote Originally Posted by Barthayn View Post
    Thanks, I had you third last step and see sec^2y and I was all like "what do I do? Help someone!". Thanks for pointing out the unit trig identity. It was much easier than it appeared
    Isn't this always the case?
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