How Do You Solve This Partial Derivatives Problem?

[Questions999]

I have z=(e^y)φ*[y*e^(x^2/2y^2)].I have to prove that y*(dz/dx) -x*(dz/dy)=0.First of all what does φ mean there?

 

[Fredrik]

I have z=(e^y)\varphi*[y*e^(x^2/2y^2)].I have to prove that y*(dz/dx) -x*(dz/dy)=0.First of all what does φ mean there?

Are you sure that this isn't specified in the problem or earlier in the text? My first guess is that it's just some number and that it won't contribute to the final result. But the only way I can verify that is to assume that it is, and then solve the problem under that assumption. This seems like something you should try yourself. Have you tried to compute the partial derivatives of the z defined by
$$z=e^y φye^{\frac{x^2}{2y^2}}$$ where $\varphi$ is just a number?

 

[Ray Vickson]

I have z=(e^y)φ*[y*e^(x^2/2y^2)].I have to prove that y*(dz/dx) -x*(dz/dy)=0.First of all what does φ mean there?

Is e^(x^2/2y^2) supposed to be
$$e^{\left( \frac{x^2}{2y^2}\right)}$$
or is it
$$e^{\left( \frac{x^2}{2}y^2\right)} ?$$
If you mean the former, you need parentheses, like this: e^(x^2/(2y^2)); if you mean the latter, you can just leave it as is, since that is what you actually wrote!

 

[Sunil Simha]

Hello

Does the original statement read
$z=e^{y}\varphi(ye^{\frac{x^{2}}{2y^{2}}})$?

If yes $\varphi(ye^{\frac{x^{2}}{2y^{2}}})$ could be a function. If they specifically have told $\varphi$*(the stuff in the bracket), then $\varphi$ may be a constant.