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Math Help - How to do this Inverse Trig Function?

  1. #1
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    How to do this Inverse Trig Function?

    Find the derivative of y
    y=sin^-1 3/t^2
    All I know is the derivative of sin^-1 is 1/square root of 1-x^2
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  2. #2
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    You know that: \frac{d}{dx} \sin^{-1} ({\color{red}u}) = \frac{1}{\sqrt{1-{\color{red}u}^2}} \cdot ({\color{red}u})'

    Making the analogy: y' = \frac{1}{\sqrt{1 - \left({\color{red}\frac{3}{t^2}}\right)^2}} \cdot \left({\color{red}\frac{3}{t^2}}\right)'
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JimDavid View Post
    All I know is the derivative of sin^-1 is 1/square root of 1-x^2
    Im assuming you are asking to find \frac{d}{dx}\bigg[\arcsin\left(\frac{3}{x^2}\right)\bigg]?

    You have two options...you could use the fact that \arcsin\left(\frac{1}{x}\right)=\csc^{-1}\left(x\right) to eliminate the problem negative exponent...this is assuming you know that derivative...or just use the chain rule

    \frac{d}{dx}\bigg[\arcsin\left(\frac{3}{x^2}\right)\bigg]=\frac{1}{\sqrt{1-\left(\frac{3}{x^2}\right)^2}}\cdot\left[\frac{3}{x^2}\right]'

    EDIT: The war begins anew...
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    Im assuming you are asking to find \frac{d}{dx}\bigg[\arcsin\left(\frac{3}{x^2}\right)\bigg]?

    You have two options...you could use the fact that \arcsin\left(\frac{1}{x}\right)=\csc^{-1}\left(x\right) to eliminate the problem negative exponent...this is assuming you know that derivative...or just use the chain rule

    \frac{d}{dx}\bigg[\arcsin\left(\frac{3}{x^2}\right)\bigg]=\frac{1}{\sqrt{1-\left(\frac{3}{x^2}\right)^2}}\cdot\left[\frac{3}{x^2}\right]'

    EDIT: The war begins anew...
    I haven't learn about arcsin yet, just the basic 6 tri functions.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JimDavid View Post
    I haven't learn about arcsin yet, just the basic 6 tri functions.
    \arcsin(x)=sin^{-1}(x)
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