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Math Help - deriv again

  1. #1
    Junior Member
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    deriv again

    I don't know what I did wrong...

    Find y' of y=cos^-1(e^2x)

    Ok, so cos^-1 is sec.

    Then I got sec(e^2x)tan(e^2x)(2e^2x)(2).

    Not sure if the (2e^2x)(2) part is right because the online hw keep telling me I'm wrong.
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Mmm... be careful with the notation.

    \cos^{-1}x=\arccos x

    --

    You wanna compute y=\sec e^{2x}

    You know that (\sec u)'=u'\cdot\sec u\tan u
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  3. #3
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    So then I got sec(e^2x)tan(e^2x)(2e^2x).

    I'm still not getting it right.
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  4. #4
    Eater of Worlds
    galactus's Avatar
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    You want to find the derivative of y=cos^{-1}(e^{2x})?.

    Chain rule.

    The derivative of cos^{-1}(x)=\frac{-1}{\sqrt{1-x^{2}}}

    So, the derivative of cos^{-1}(e^{2x})=\frac{-1}{\sqrt{1-e^{4x}}}\cdot{\underbrace{2e^{2x}}_{\text{chain rule}}}

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