- #1

- 42

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter jaguar7
- Start date

- #1

- 42

- 0

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 967

[tex]\frac{x^2}{4}+ y^2+ z^2= 3[/tex]

is not a function, it is an equation. If you mean that equation defines z implicitely as a function of x and y, then, differentiating with respect to x, [itex]x/2+ 2z z_x= 0[/itex] so that [itex]z_x= -x/4z[/itex] and [itex]2y+ 2z z_y= 0[/itex] so that [itex]z_y= -y/z[/itex]. Notice that, in the first case, the derivative of y with respect to x is 0 and, in the second, the derivative of x with resepect to y is 0. That is because they are independent variables.

Of course, we could just as easily think of that equation as defining y in terms of the variables x and z and have [itex]x/2+ 2yy_x= 0[/itex] and [itex]2y y_z+ 2z= 0[/itex] or we could think of it as defining x in terms of y and z so that [itex](x/2)x_y+ 2y= 0[/itex] and [itex](x/2)x_z+ 2z= 0[/itex].

Share: