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Math Help - derivative 1

  1. #1
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    derivative 1

    y= tan^3 (sqrt(4 - x^4))

    i did
    y = [tan(sqrt4 - x^4)]^3
    y'= 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[(4-x^4)^-1/2]'
    = 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2](4x^3)
    = 12x^3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2]

    is this the final answer?

    do i have to do something with [1/2(4-x^4)^-1/2] ?
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  2. #2
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    Quote Originally Posted by haebinpark View Post
    y= tan^3 (sqrt(4 - x^4))

    i did
    y = [tan(sqrt4 - x^4)]^3
    y'= 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[(4-x^4)^-1/2]'
    = 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2](4x^3)
    = 12x^3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2]

    That isn't the correct derivative.

    is this the final answer?

    do i have to do something with [1/2(4-x^4)^-1/2] ?
    y'=3u^2\frac{du}{dx} where u=tan(\sqrt{4-x^4})

    \frac{du}{dx}=sec^2(\sqrt{4-x^4})\frac{dv}{dx} where v=\sqrt{4-x^4}

    \frac{dv}{dx}=\frac{4x^3}{2\sqrt{4-x^4}}=\frac{2x^3}{\sqrt{4-x^4}}

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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by haebinpark View Post
    y= tan^3 (sqrt(4 - x^4))

    i did
    y = [tan(sqrt4 - x^4)]^3
    y'= 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[(4-x^4)^-1/2]'
    = 3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2](4x^3)
    = 12x^3[tan(sqrt4 - x^4)]^2 * [sec^2 (sqrt(sqrt4 - x^4))]*[1/2(4-x^4)^-1/2]

    is this the final answer?

    do i have to do something with [1/2(4-x^4)^-1/2] ?
    Recall the chain rule
    \frac{d}{dx}\left[\tan^3\sqrt{4-x^4}\right]=3\tan^2\sqrt{4-x^4}\cdot\sec^2\sqrt{4-x^4}\cdot\frac{1}{2}(4-x^4)^{-1/2}(-4x^3)

    =\frac{-6x^3\tan^2\sqrt{4-x^4}\cdot\sec^2\sqrt{4-x^4}}{\sqrt{4-x^4}}
    Last edited by VonNemo19; October 28th 2009 at 12:37 PM.
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