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Math Help - Need help, quickly. Implicit differtiation

  1. #1
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    Question Need help, quickly. Implicit differtiation

    I have been having trouble with this question. A few people have helped, but I've gotten no where from that. please help!

    Question: Show that for the relation ((x/y)^.5) + ((y/x)^.5) = 10, x does not equal y which does not equal 0, dy/dx = y/x

    Thanks
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  2. #2
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    Hello, elle1019!

    This is a messy one . . .


    Show that the relation: . \left(\frac{x}{y}\right)^{\frac{1}{2}} + \left(\frac{y}{x}\right)^{\frac{1}{2}} \:= \;10

    \text{ where }x \neq y\,\text{ and }\,x,y \neq 0,\;\text{ produces: }\;\frac{dy}{dx} \:=\: \frac{y}{x}
    Differentiate:

    . . \frac{1}{2}\left(\frac{x}{y}\right)^{-\frac{1}{2}}\left(\frac{y - x\frac{dy}{dx}}{y^2}\right) + \frac{1}{2}\left(\frac{y}{x}\right)^{-\frac{1}{2}}\left(\frac{x\frac{dy}{dx} - y}{x^2}\right) \;=\;0

    . . . . . . . . \frac{1}{2}\cdot\frac{y^{\frac{1}{2}}}{x^{\frac{1}  {2}}}\cdot\frac{y-x\frac{dy}{dx}}{y^2} + \frac{1}{2}\cdot\frac{x^{\frac{1}{2}}}{y^{\frac{1}  {2}}}\cdot\frac{x\frac{dy}{dx}-y}{x^2} \;=\;0

    . . . . . . . . . . . . . . . . . \frac{y-x\frac{dy}{dx}}{2x^{\frac{1}{2}}y^{\frac{3}{2}}} + \frac{x\frac{dy}{dx}-y}{2x^{\frac{3}{2}}y^{\frac{1}{2}}} \;=\;0


    Multiply by 2x^{\frac{3}{2}}y^{\frac{3}{2}}\!:\;\;x\left(y-x\frac{dy}{dx}\right) + y\left(x\frac{dy}{dx} - y\right) \;=\;0

    . . . . . . . . . . . . . . . . . xy - x^2\frac{dy}{dx} + xy\frac{dy}{dx} - y^2 \;=\;0

    . . . . . . . . . . . . . . . . . . . . . . . xy\frac{dy}{dx} - x^2\frac{dy}{dx} \;=\;y^2 - xy

    . . . . . . . . . . . . . . . . . . . . . . . . x(y-x)\frac{dy}{dx} \;=\;y(y-x)

    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \frac{dy}{dx} \;=\;\frac{y(y-x)}{x(y-x)}
    . . Therefore: . \frac{dy}{dx} \:=\:\frac{y}{x}

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