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Math Help - Help on derivative

  1. #1
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    Help on derivative

    1. tan\frac{x}{y}=x+y
    2. \sqrt{xy}=1+x^2y
    I know that you use implicit differentiation, but don't know when to use the chain rule or the product rule.

    These are the answers in my book:

    1. y' = \frac{ysec^2(\frac{x}{y})-y^2}{y^2+xsec^2(\frac{x}{y})}
    2. y' = \frac{4xy\sqrt{xy}-y}{x-2x^2\sqrt{xy}}
    Last edited by c_323_h; June 18th 2006 at 11:17 AM.
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  2. #2
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    Quote Originally Posted by c_323_h
    1. tan\frac{x}{y}=x+y
    Take derivative,
    (x/y)'\sec^2 (x/y)=1+y'
    Thus,
    \frac{(x)'y-x(y)'}{y^2}\cdot \sec^2(x/y)=1+y'
    Thus,
    \frac{y-xy'}{y^2}\cdot \sec^2(x/y)=1+y'Thus,
    y-xy'=y^2\cos^2(x/y)+y^2\cos^2(x/y)y'
    Thus,
    y'(-x-y^2\cos^2(x/y))=y^2\cos^2(x/y)-y
    Thus,
    y'=\frac{y-y^2\cos^2(x/y)}{x+y^2\cos^2(x/y)}
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  3. #3
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    Quote Originally Posted by c_323_h
    2. \sqrt{xy}=1+x^2y
    \sqrt{xy}=1+x^2y

    \frac{1}{2}\frac{1}{\sqrt{xy}}(y+xy')=2xy'

    \frac{1}{2}\frac{1}{\sqrt{xy}}xy'-2xy'=-\frac{1}{2}\frac{1}{\sqrt{xy}}y

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

    y' = -\frac{\frac{1}{2}\sqrt{\frac{y}{x}}}{\frac{1}{2} \sqrt{\frac{x}{y}} - 2x}

    y' = -\frac{\frac{1}{2}\sqrt{\frac{y}{x}}}{\frac{1}{2} \sqrt{\frac{x}{y}} - 2x}*\frac{2\sqrt{xy}}{2\sqrt{xy}} = \frac{y}{x-4x \sqrt{xy}}=\frac{y}{x}\frac{1}{1-4 \sqrt{xy}}

    -Dan
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