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Thread: 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.

    $\displaystyle
    \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)
    $
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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.

    $\displaystyle
    \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)
    $
    let $\displaystyle y = \sin^{-1}(u)$ , rewrite as the inverse function $\displaystyle \sin{y} = u$

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

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

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

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

    $\displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$
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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.

    $\displaystyle
    \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)
    $
    Since you have a composition of functions, you would use the chain rule.

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


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


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


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

    $\displaystyle y = \arcsin{u}$

    $\displaystyle \sin{y} = u$

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

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

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

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

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



    Now, from the Pythagorean Identity, we have

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

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


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

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



    Therefore:

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

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

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

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

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

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

    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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    Quote Originally Posted by DBA View Post
    Thanks for your answer.
    How do I know that $\displaystyle \cos{y} = \sqrt{1-u^2}$

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

    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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