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Math Help - Derivatives of inverse trigonometric functions

  1. #1
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    Derivatives of inverse trigonometric functions

    Given that y=arcsinx, show that

    (1-x^2)d^2y/dx^2 - x(dy/dx)=0
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  2. #2
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    \frac{dy}{dx}=\frac1{\sqrt{1-x^2}} whereat \frac{d^2y}{dx^2}=\cdots

    Then just make the proper substitutions to prove it.
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  3. #3
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    Thanks for replying, but could you please elaborate further..
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  4. #4
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    You're expected to know that \frac{d^2y}{dx^2} is the second derivative, whereas the first is \frac1{\sqrt{1-x^2}} then by just knowing this you can compute \frac{d^2y}{dx^2}. Once you get \frac{dy}{dx} and \frac{d^2y}{dx^2} put them together into the equation and the rest is algebra.
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  5. #5
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    d2y/dx2 = (1-x^2)^-3/2...? I can't sub in dy/dx and d2y/dx2 and make it equal 0
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  6. #6
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    \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}=\left( 1-x^{2} \right)^{-1/2}\implies \frac{d^{2}y}{dx^{2}}=-\frac{1}{2}\cdot (-2x)\left( 1-x^{2} \right)^{-3/2}, whereat \frac{d^{2}y}{dx^{2}}=\frac{x}{\left( 1-x^{2} \right)^{3/2}}.

    Dinner is served.
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  7. #7
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    OH man. I completely ignored the x in 2x. n00b error. Thank you for your help.
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  8. #8
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    Hello, Erghhh!

    It seems straight-forward enough.
    Exactly where is your difficulty?


    Given that: . y\:=\:\arcsin x,

    show that: . (1-x^2)\,\frac{d^2y}{dx^2} - x\,\frac{dy}{dx} \:=\:0

    We have: . \frac{dy}{dx} \:=\:\frac{1}{\sqrt{1-x^2}} \:=\:(1-x^2)^{-\frac{1}{2}}

    And: . \frac{d^2y}{dx^2} \;=\;-\frac{1}{2}(1-x^2)^{-\frac{3}{2}} (-2x) \;=\;\frac{x}{(1-x^2)^{\frac{3}{2}}}


    Then: . (1-x^2)\;\;\cdot\;\;\frac{d^2y}{dx^2} \;\;-\;\;x\;\;\cdot\;\;\frac{dy}{dx}

    . . . = \;(1-x^2)\cdot\frac{x}{(1-x^2)^{\frac{3}{2}}} - x\cdot\frac{1}{(1-x^2)^{\frac{1}{2}}}

    . . . = \;\frac{x}{(1-x^2)^{\frac{1}{2}}} - \frac{x}{(1-x^2)^{\frac{1}{2}}}


    . . . =\;0

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  9. #9
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    Thanks, i see what i'm doing now. i just made a stupid error.
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