Partial Differentiation

**Schniz2** · May 25th 2008, 09:22 PM

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

**Reckoner** · May 25th 2008, 09:45 PM

Originally Posted by Schniz2

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

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

**Chris L T521** · May 25th 2008, 10:21 PM

Originally Posted by Schniz2

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 $\frac{\partial f}{\partial y} \ \text{and} \ \frac{\partial f}{\partial x}$ .

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

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

$<br /> \begin{aligned} \therefore\frac{dy}{dx}&=-\frac{2xy-\frac{y}{2\sqrt{xy}}}{x^2-\frac{x}{2\sqrt{xy}}}\\<br /> &=-\frac{4(xy)^{\frac{3}{2}}-y}{2x^2\sqrt{xy}-x} \\<br /> &=\color{red}\boxed{\frac{y-4(xy)^{\frac{3}{2}}}{x-2x^2\sqrt{xy}}}<br /> \end{aligned}$

**Schniz2** · May 25th 2008, 11:59 PM

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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