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Math Help - Partial Differentiation

  1. #1
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    Question Partial Differentiation

    I've tried this one tonnes of times... and i keep getting the same incorrect solution... the sol'n is down the bottom in red..

    not sure if im differentiating incorrectly or if my answer is just in a different form :S

    Equation 6 is the equation typed on the RHS.

    Sorry for the messy writing :P. it would take me about 2 hours to type it up lol.

    Thanks
    Attached Thumbnails Attached Thumbnails Partial Differentiation-eqtn-6-copy.jpg  
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Schniz2 View Post
    I've tried this one tonnes of times... and i keep getting the same incorrect solution...
    You should be more careful then:

    \frac\partial{\partial x}\left[\sqrt{xy}\right] = \frac\partial{\partial x}\left[(xy)^{1/2}\right] = \frac12{\color{red}y}(xy)^{-1/2} = \frac{\color{red}y}{2\sqrt{xy}}

    and

    \frac\partial{\partial y}\left[\sqrt{xy}\right] = \frac\partial{\partial y}\left[(xy)^{1/2}\right] = \frac12{\color{red}x}(xy)^{-1/2} = \frac{\color{red}x}{2\sqrt{xy}}
    Last edited by Reckoner; May 25th 2008 at 09:49 PM. Reason: Added intermediate steps
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Schniz2 View Post
    I've tried this one tonnes of times... and i keep getting the same incorrect solution... the sol'n is down the bottom in red..

    not sure if im differentiating incorrectly or if my answer is just in a different form :S

    Equation 6 is the equation typed on the RHS.

    Sorry for the messy writing :P. it would take me about 2 hours to type it up lol.

    Thanks
    \sqrt{xy}=1+x^2y \implies x^2y-\sqrt{xy}+1=0

    Letting f(x,y)=x^2y-\sqrt{xy}+1, we then can find \frac{\partial f}{\partial y} \ \text{and} \ \frac{\partial f}{\partial x}.

    \frac{\partial f}{\partial y}=x^2-\frac{x}{2\sqrt{xy}}

    \frac{\partial f}{\partial x}=2xy-\frac{y}{2\sqrt{xy}}

    <br />
\begin{aligned} \therefore\frac{dy}{dx}&=-\frac{2xy-\frac{y}{2\sqrt{xy}}}{x^2-\frac{x}{2\sqrt{xy}}}\\<br />
&=-\frac{4(xy)^{\frac{3}{2}}-y}{2x^2\sqrt{xy}-x} \\<br />
&=\color{red}\boxed{\frac{y-4(xy)^{\frac{3}{2}}}{x-2x^2\sqrt{xy}}}<br />
\end{aligned}
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  4. #4
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    ahh, thanks guys. i've been doing them all day and i seem to be making tonnes of stupid little mistakes.

    thanks again
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