Question

Find the four second partial derivatives. Observe that the second mixed partials are equal. $$ z=x^{2}-2 x y+3 y^{2} $$

Short answer

\(\frac{{\partial}^2 z}{{\partial} x^2} = 2\), \(\frac{{\partial}^2 z}{{\partial} y^2} = 6\), \(\frac{{\partial}^2 z}{{\partial} y \partial} x = -2\), \(\frac{{\partial}^2 z}{{\partial} x \partial} y = -2\). Yes, the second mixed partials are equal.

Step-by-step solution

Find the first order partial derivatives

Differentiate \(z = x^2 - 2xy + 3y^2\) partially with respect to \(x\) and then \(y\) to get first order partial derivatives. \(\frac{{\partial} z}{{\partial} x} = 2x - 2y\) \(\frac{{\partial} z}{{\partial} y} = -2x + 6y\)

Find the second order partial derivatives

Differentiate \(\frac{{\partial} z}{{\partial} x}\) and \(\frac{{\partial} z}{{\partial} y}\) again partially with respect to \(x\) and \(y\) respectively, to find \(\frac{{\partial}^2 z}{{\partial} x^2}\) and \(\frac{{\partial}^2 z}{{\partial} y^2}\). \(\frac{{\partial}^2 z}{{\partial} x^2} = 2\) \(\frac{{\partial}^2 z}{{\partial} y^2} = 6\)

Find the mixed second order partial derivatives

Differentiate \(\frac{{\partial} z}{{\partial} x}\) with respect to \(y\) and \(\frac{{\partial} z}{{\partial} y}\) with respect to \(x\) to find the mixed second order partial derivatives. \(\frac{{\partial}^2 z}{{\partial} y \partial} x = -2\) \(\frac{{\partial}^2 z}{{\partial} x \partial} y = -2\)

Verify Equality of Mixed Partial Derivatives

Check if \(\frac{{\partial}^2 z}{{\partial} y \partial} x = \frac{{\partial}^2 z}{{\partial} x \partial} y\). Here, both are equal to -2, so we can indeed confirm that they are equal.