How to take the Partial Derivatives of a Function that is Defined Implicitly?

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SUMMARY

The discussion focuses on taking partial derivatives of a function defined implicitly, specifically using the equation x²/4 + y² + z² = 3. It clarifies that this equation does not represent a function but defines z implicitly as a function of x and y. The partial derivatives are derived as zₓ = -x/(4z) and zᵧ = -y/z, highlighting that the derivatives of independent variables x and y with respect to each other are zero. The discussion also explores alternative interpretations of the equation defining y or x in terms of the other variables.

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How does one take the partial derivatives of a function that is defined implicitly? For example, the function, x^2 / 4 + y^2 + z^2 = 3.
 
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To take the partial derivatives of a function, you first have to specify the function!
[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].
 

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