find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ as a functions of $x, y,\text{ and }z$ assuming that $z=f(x,y)$ satisfies the equation
$$x^{2/3}+y^{2/3}+z^{2/3}=1$$

Question

find $\frac{\partial z}{\partial x}$ and  $\frac{\partial z}{\partial y}$ as a functions of $x, y,\text{ and }z$ assuming that $z=f(x,y)$ satisfies the equation
$$x^{2/3}+y^{2/3}+z^{2/3}=1$$

anonymous · Answer

Taking a partial derivative.

For example:

$$\frac{\partial z}{ \partial y}$$ means that you derive with respect to y and threat x,z as a constant.

richyw · Answer

yeah I know that. but I don't really get what this question is asking. there has got to be the chain rule in there.

klimenkov · Answer

All you need is$$\frac {\partial  }{\partial x}x^n=n\cdot x^{n-1}$$try to do this with n=2/3.

anonymous · Answer

for example if u want to get   $\huge \frac{\partial z}{\partial x} $ u have    $\huge  x^{-\frac{1}{3}}+\huge  z^{-\frac{1}{3}}\frac{\partial z}{\partial x}=0 $ 
and now u just need to derive $\huge  z^{-\frac{1}{3}}$  from original equation and put it in the equation above

klimenkov · Answer

Example: $x+y+z=1$. Find $\frac {\partial z}{\partial y}$ when z=f(x,y). You have to derive both parts of equation respect to y:$$\frac {\partial x}{\partial y}+\frac {\partial y}{\partial y}+\frac {\partial z}{\partial y}=\frac {\partial }{\partial y}1$$$$0+1+\frac {\partial z}{\partial y}=0$$$$\frac {\partial z}{\partial y}=-1$$Just make the same thing.