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Math Help - differentiation

  1. #1
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    differentiation

    Hey there, I've got a question thats been bugging me for ages...I have no idea how to start it!
    Differentiate tan(arcsin (1-x^2)^1/2)

    Any help will be appreciated, thanks!
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: differentiation

    Do you know the chain rule?
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  3. #3
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    Re: differentiation

    yes, but I dont see how I can use that here...
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  4. #4
    Super Member Quacky's Avatar
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    Re: differentiation

    Let u=\arcsin{(1-x^2)}, and then, if necessary, let v=1-x^2 and use the chain rule twice.
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  5. #5
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    Re: differentiation

    So I would get something like f'(x)sec^2(f(x)) with f(x) being arcsin(1-x^2)^1/2 ?
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  6. #6
    Super Member Quacky's Avatar
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    Re: differentiation

    Quote Originally Posted by maths13 View Post
    So I would get something like f'(x)sec^2(f(x)) with f(x) being arcsin(1-x^2)^1/2 ?
    Yeah.
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: differentiation

    Another approach might be to use:

    \tan\left(\arcsin\left(\sqrt{1-x^2}\right)\right)=\sqrt{x^{-2}-1}

    Then use the chain rule from there...
    Last edited by MarkFL; December 24th 2011 at 11:06 AM.
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  8. #8
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    Re: differentiation

    Quote Originally Posted by maths13 View Post
    Hey there, I've got a question thats been bugging me for ages...I have no idea how to start it!
    Differentiate tan(arcsin (1-x^2)^1/2)

    Any help will be appreciated, thanks!
    An alternative is to note that

    \displaystyle \begin{align*} \tan{\theta} &\equiv \frac{\sin{\theta}}{\cos{\theta}} \\ &\equiv \pm \frac{\sin{\theta}}{\sqrt{1 - \sin^2{\theta}}} \end{align*}

    Therefore

    \displaystyle \begin{align*} \tan{\left[\arcsin{\left(\sqrt{1 - x^2}\right)}\right]} &\equiv \pm \frac{\sin{\left[\arcsin{\left(\sqrt{1 - x^2}\right)}\right]}}{\sqrt{1 - \sin^2{\left[\arcsin{\left(\sqrt{1 - x^2}\right)}\right]}}} \\ &\equiv \pm \frac{\sqrt{1 - x^2}}{\sqrt{1 - \left(\sqrt{1 - x^2}\right)^2}} \\ &\equiv \pm \frac{\sqrt{1 - x^2}}{\sqrt{1 - \left(1 - x^2\right)}} \\ &\equiv \pm \frac{\sqrt{1 - x^2}}{\sqrt{x^2}} \\ &\equiv \pm \sqrt{\frac{1 - x^2}{x^2}} \\ &\equiv \pm \left(x^{-2} - 1 \right)^{\frac{1}{2}}\end{align*}


    This should be much easier to differentiate...
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