Calculus problem need assistance

[Hira Javaid]

can someone check this and tell me if i am doing is write or wrong and if its right then what in the mistake when i am taking partial derivative of 2nd equation because i am getting infinity after solving this..

 

[JeffM]

I must be missing something. We are told that x, y, and z are functions of u and v, but presumably u and v are independent of each other. So why do we say that

$$\left (\dfrac{\delta}{\delta v} x^3z \right ) = 0.$$

 

[Hira Javaid]

I must be missing something. We are told that x, y, and z are functions of u and v, but presumably u and v are independent of each other. So why do we say that

$$\left (\dfrac{\delta}{\delta v} x^3z \right ) = 0.$$

I treated x an z as constants as we have to (fibd)find the derivative of u edited.....fibd to find

 

[Deleted member 4993]

I treated x an z as constants as we have to (fibd)find the derivative of u edited.....fibd to find

5 th line of tells us that "... x, y and z as a function of u and v"

So x, y and z CANNOT be treated as constants....[edited]

 

[JeffM]

I think I might try working with the second equation to get

$$y = 1 + \frac{1}{2} * (uv - x^3z) \implies \\ \dfrac{\delta y}{\delta v} = \frac{1}{2}* \left ( u - x^3 * \dfrac{\delta z}{\delta v} - 3x^2z * \dfrac{\delta x}{\delta v} \right ) \text { and } \dfrac{\delta y}{\delta u} = \frac{1}{2} * \left ( v - x^3 * \dfrac{\delta z}{\delta u} - 3x^2z * \dfrac{\delta x}{\delta u} \right ).$$
Then, with a bunch of probably nasty algebra, you can eliminate y from the first and third equations. Now you have only two dependent variables. If you can eliminate either of them (for example z), you are down to a single equation relating two independent variable to one dependent variable.

There may of course be a much more elegant attack,