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Math Help - Prove formula for derivative inverse trig function

  1. #1
    DBA
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    Prove formula for derivative inverse trig function

    I need to prove the following. I do not know how to do that. Thanks for any help.

    <br />
\frac{d}{dx} \left[\sin^-1\left(u \left( x\right)\right)\right] = \frac{1}{\sqrt[2]{1- \left[ u \left(x\right)\right]^2}}  \frac{d}{dx} u\left(x\right)<br />
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    Quote Originally Posted by DBA View Post
    I need to prove the following. I do not know how to do that. Thanks for any help.

    <br />
\frac{d}{dx} \left[\sin^-1\left(u \left( x\right)\right)\right] = \frac{1}{\sqrt[2]{1- \left[ u \left(x\right)\right]^2}}  \frac{d}{dx} u\left(x\right)<br />
    let y = \sin^{-1}(u) , rewrite as the inverse function \sin{y} = u

    \frac{d}{dx}\left[\sin{y} = u\right]

    \cos{y} \cdot \frac{dy}{dx} = \frac{du}{dx}

    since \sin{y} = u , \cos{y} = \sqrt{1-u^2}

    \frac{dy}{dx} = \frac{1}{\cos{y}} \cdot \frac{du}{dx}

    \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}
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  3. #3
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    Quote Originally Posted by DBA View Post
    I need to prove the following. I do not know how to do that. Thanks for any help.

    <br />
\frac{d}{dx} \left[\sin^-1\left(u \left( x\right)\right)\right] = \frac{1}{\sqrt[2]{1- \left[ u \left(x\right)\right]^2}}  \frac{d}{dx} u\left(x\right)<br />
    Since you have a composition of functions, you would use the chain rule.

    y = \arcsin{(u(x))}.


    \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}.


    Clearly, \frac{du}{dx} = \frac{d}{dx}\left[u(x)\right].


    To find \frac{dy}{du}, we have

    y = \arcsin{u}

    \sin{y} = u

    \frac{d}{du}(\sin{y}) = \frac{d}{du}(u)

    \frac{dy}{du}\,\frac{d}{dy}(\sin{y}) = 1

    \cos{y}\,\frac{dy}{du} = 1

    \frac{dy}{du} = \frac{1}{\cos{y}}

    \frac{dy}{du} = \frac{1}{\cos{\arcsin{u}}}.



    Now, from the Pythagorean Identity, we have

    \sin^2{x} + \cos^2{x} = 1

    \cos{x} = \sqrt{1 - \sin^2{x}}


    So \frac{dy}{du} = \frac{1}{\sqrt{1 - (\sin{\arcsin{u}})^2}}

     = \frac{1}{\sqrt{1 - u^2}}.



    Therefore:

    \frac{dy}{dx} = \frac{1}{\sqrt{1 - [u(x)]^2}}\,\frac{d}{dx}[u(x)].
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  4. #4
    DBA
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    Quote Originally Posted by skeeter View Post
    let y = \sin^{-1}(u) , rewrite as the inverse function \sin{y} = u

    \frac{d}{dx}\left[\sin{y} = u\right]

    \cos{y} \cdot \frac{dy}{dx} = \frac{du}{dx}

    since \sin{y} = u , \cos{y} = \sqrt{1-u^2}

    \frac{dy}{dx} = \frac{1}{\cos{y}} \cdot \frac{du}{dx}

    \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}
    Thanks for your answer.
    How do I know that \cos{y} = \sqrt{1-u^2}

    And I get confused with the u(x). Now we just use u. How do I end up with u(x) again. I do not get that at all.
    Would you mind explaining that a little more?
    Thanks
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  5. #5
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    Quote Originally Posted by DBA View Post
    Thanks for your answer.
    How do I know that \cos{y} = \sqrt{1-u^2}

    because \textcolor{red}{\cos^2{y} + \sin^2{y} = 1}

    And I get confused with the u(x). Now we just use u. How do I end up with u(x) again. I do not get that at all.
    Would you mind explaining that a little more?

    u(x) just means that u is a function of x ... writing u by itself means the same thing.
    ...
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  6. #6
    DBA
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    Quote Originally Posted by skeeter View Post
    ...
    Oh, ok. I got it now. Thank you!
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