Thread: Partial Differentiation

1. 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

2. 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:

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

and

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

3. 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
$\displaystyle \sqrt{xy}=1+x^2y \implies x^2y-\sqrt{xy}+1=0$

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

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

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

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

4. ahh, thanks guys. i've been doing them all day and i seem to be making tonnes of stupid little mistakes.

thanks again