Thread: Partial diff. is this correct

1. Partial diff. is this correct

Hi everyone,

Need help to verify this working.Thank you for all help & support.

Find $\displaystyle \frac{dz}{dx}$ & $\displaystyle \frac{dz}{dy}$ for
$\displaystyle x^2y^3z+yz^2-xy=0$
$\displaystyle 2xy^3z+x^2y^3\frac{dz}{dx}+2yz\frac{dz}{dx}-y=0$
$\displaystyle (x^2y^3+2yz)\frac{dz}{dx}=y-2xy^3z$
$\displaystyle \frac{dz}{dx}=\frac{y-2xy^3z}{x^2y^3+2yz}$

$\displaystyle 3x^2y^2z+x^2y^3\frac{dz}{dy}+z^2+2yz\frac{dz}{dy}-x=0$
$\displaystyle \frac{dz}{dy}=\frac{x-3x^2y^2z-z^2}{x^2y^3+2yz}$

Thank you in advance for all help & support,really appreciate.

2. What is the right hand side of this function? Is it set $\displaystyle =0$?

3. Hi ProveIt,

yes, it's equal to 0.

4. $\displaystyle x^2y^3z + yz^2 - xy = 0$

$\displaystyle x^2y^3z + yz^2 = xy$

$\displaystyle z^2 + x^2y^2z = x$

$\displaystyle z^2 + x^2y^2z + \left(\frac{1}{2}x^2y^2\right)^2 = \left(\frac{1}{2}x^2y^2\right)^2 + x$

$\displaystyle \left(z + \frac{1}{2}x^2y^2\right)^2 = \frac{1}{4}x^4y^4 + x$

$\displaystyle z + \frac{1}{2}x^2y^2 = \pm \sqrt{\frac{1}{4}x^4y^4 + x}$

$\displaystyle z + \frac{1}{2}x^2y^2 = \pm \sqrt{\frac{x^4y^4 + 4x}{4}}$

$\displaystyle z + \frac{1}{2}x^2y^2 = \pm \frac{\sqrt{x^4y^4 + 4x}}{2}$

$\displaystyle z = \frac{-x^2y^2 \pm \sqrt{x^4y^4 + 4x}}{2}$.

Now try to differentiate with respect to $\displaystyle x$ and $\displaystyle y$.

5. thank you for replying.there is a different alternative to answer this question though.