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Math Help - Differentiation?

  1. #1
    Super Member fardeen_gen's Avatar
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    Differentiation?

    If u = e^(arcsin (y/u)), then prove that d^2y/dx^2 = 2(x^2 + y^2)/(x - y)^3 , where u = √(x^2 + y^2)
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by fardeen_gen View Post
    If u = e^(arcsin (y/u)), then prove that d^2y/dx^2 = 2(x^2 + y^2)/(x - y)^3 , where u = √(x^2 + y^2)
    This is equivalent to \ln(u)=\arcsin(y/u)

    -> Implicit differentiation with respect to x...

    Left side :
    \frac{\partial}{\partial x} (\ln(u))=\frac{\partial u}{\partial x} \cdot \frac 1u=\left(\frac{\partial}{\partial x} (\sqrt{x^2+y^2})\right) \cdot \frac 1u=\frac{x+yy'}{\sqrt{x^2+y^2}}\cdot \frac 1u=\boxed{\frac{x+yy'}{u^2}}, where y'=\frac{dy}{dx}

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Right side :

    \frac{\partial}{\partial x} (\arcsin(y/u)) :
    By the chain rule, this is :
    \frac{\partial}{\partial x} (y/u) \cdot \frac{1}{\sqrt{1-(y/u)^2}}

    - \sqrt{1-(y/u)^2}=\sqrt{1-\frac{y^2}{x^2+y^2}}=\sqrt{\frac{x^2}{x^2+y^2}}=\f  rac{x}{u} (we'll assume x>0...)
    - \frac{\partial}{\partial x} (y/u)=\frac{y'u-y \cdot \frac{\partial u}{\partial x}}{u^2}
    =\frac{y'u-y \cdot \frac{x+yy'}{u}}{u^2}

    \Rightarrow \frac{\partial}{\partial x} (\arcsin(y/u))=\frac{y'u-y \cdot \frac{x+yy'}{u}}{u^2} \cdot \frac ux=\frac{y'u^2-y(x+yy')}{xu^2}

    =\frac{y'x^2+y'y^2-yx-y'y^2}{xu^2}=\frac{y'x^2-yx}{xu^2}=\boxed{\frac{y'x-y}{u^2}}
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    So we finally have :

    \frac{x+yy'}{u^2}=\frac{y'x-y}{u^2}


    That is \boxed{x+yy'=xy'-y}
    (this implies that y'=\frac{x+y}{x-y} - this will be useful for later)


    Differentiate implicitely, again with respect to x.
    1+y'^2+yy''=y'+xy''-y'=xy'', where y''=\frac{d^2y}{dx^2}

    --> y''=\frac{1+y'^2}{x-y}

    y''=\frac{1+\left(\frac{x+y}{x-y}\right)^2}{x-y}

    y''=\frac{(x-y)^2+(x+y)^2}{(x-y)^3}

    \boxed{y''=\frac{2(x^2+y^2)}{(x-y)^3}}




    (I don't know if there is an easier way to do it though...)
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