Calculations of the first, second and higher order partial derivatives are presented with examples and their solutions including detailed steps of computations. Clairaut's theorem on the equality of mixed partials under certain continuity conditions is verified through examples.

The second and higher order partial derivatives are essentially the derivatives of functions of many variables taken multiple times with respect to one or more variables.

With respect to \(x\): \(\dfrac{\partial f}{\partial x}\)

With respect to \(y\): \(\dfrac{\partial f}{\partial y}\)

With respect to \(x\) twice: \(\dfrac{\partial^2 f}{\partial x^2}\)

With respect to \(y\) twice: \(\dfrac{\partial^2 f}{\partial y^2}\)

With respect to \(x\) and then \(y\): \(\dfrac{\partial^2 f}{\partial y \partial x}\)

With respect to \(y\) and then \(x\): \(\dfrac{\partial^2 f}{\partial x \partial y}\)

a. Differentiate \(f\) with respect to \(x\): \[ \dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x}(x^2y) + \dfrac{\partial}{\partial x}(3xy^2) \\\\ = 2xy + 3y^2 \] b. Differentiate \(f\) with respect to \(y\): \[ \dfrac{\partial f}{\partial y} = \dfrac{\partial}{\partial y}(x^2y) + \dfrac{\partial}{\partial y}(3xy^2) \\\\ = x^2 + 6xy \] 2. Second-Order Partial Derivatives

a. Differentiate \(\dfrac{\partial f}{\partial x}\) with respect to \(x\): \[ \dfrac{\partial^2 f}{\partial x^2} = \dfrac{\partial}{\partial x} \left(\dfrac{\partial f}{\partial x} \right) \\\\ =\dfrac{\partial}{\partial x}(2xy + 3y^2) = 2y \] b. Differentiate \(\dfrac{\partial f}{\partial y}\) with respect to \(y\): \[ \dfrac{\partial^2 f}{\partial y^2} = \dfrac{\partial}{\partial y} \left(\dfrac{\partial f}{\partial y} \right) \\\\ = \dfrac{\partial}{\partial y}(x^2 + 6xy) = 6x \] c. Differentiate \(\dfrac{\partial f}{\partial x}\) with respect to \(y\): \[ \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial }{\partial y} \left(\dfrac{\partial f}{\partial x}\right) \\\\= \dfrac{\partial}{\partial y}(2xy + 3y^2) = 2x + 6y \] d. Differentiate \(\dfrac{\partial f}{\partial y}\) with respect to \(x\) (should give the same result as step 3 due to symmetry, if the function's mixed partials are continuous): \[ \dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial }{\partial x} \left(\dfrac{\partial f}{\partial y}\right) \\\\ =\dfrac{\partial}{\partial x}(x^2 + 6xy) = 2x + 6y \] Notice that \(\dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial^2 f}{\partial x \partial y}\), which illustrates Clairaut's theorem on the equality of mixed partials under certain continuity conditions.

a. Differentiate \(g\) with respect to \(x\): \[ \dfrac{\partial g}{\partial x} = \dfrac{\partial}{\partial x}(e^{xy}) + \dfrac{\partial}{\partial x}(\sin(x)y^2) \\\\= ye^{xy} + \cos(x)y^2 \] b. Differentiate \(g\) with respect to \(y\): \[ \dfrac{\partial g}{\partial y} = \dfrac{\partial}{\partial y}(e^{xy}) + \dfrac{\partial}{\partial y}(\sin(x)y^2) \\\\= xe^{xy} + 2y\sin(x) \] 2. Second-Order Partial Derivatives

a. Differentiate \(\dfrac{\partial g}{\partial x}\) with respect to \(x\): \[ \dfrac{\partial^2 g}{\partial x^2} = \dfrac{\partial}{\partial x} \left(\dfrac{\partial g}{\partial x}\right) \\\\=\dfrac{\partial}{\partial x}(ye^{xy} + \cos(x)y^2) \\\\= y^2e^{xy} - y^2\sin(x) \] This step involves applying the product rule for differentiation to both terms separately.

b. Differentiate \(\dfrac{\partial g}{\partial y}\) with respect to \(y\): \[ \dfrac{\partial^2 g}{\partial y^2} = \dfrac{\partial}{\partial y} \left(\dfrac{\partial g}{\partial y}\right) \\\\ = \dfrac{\partial}{\partial y}(xe^{xy} + 2y\sin(x)) \\\\= x^2e^{xy} + 2\sin(x) \] Again, the product rule is applied, this time focusing on how the exponential term and the sine term change with respect to \(y\).

c. Differentiate \(\dfrac{\partial g}{\partial x}\) with respect to \(y\): \[ \dfrac{\partial^2 g}{\partial x \partial y} = \dfrac{\partial }{\partial y} \left(\dfrac{\partial g}{\partial x}\right) \\\\ = \dfrac{\partial}{\partial y}(ye^{xy} + \cos(x)y^2) \\\\= e^{xy} + xye^{xy} + 2y\cos(x) \] d. Differentiate \(\dfrac{\partial g}{\partial y}\) with respect to \(x\) (again, the result is the same as step 3 due to symmetry and the continuity of the mixed partials): \[ \dfrac{\partial^2 g}{\partial y \partial x} = \dfrac{\partial }{\partial x} \left(\dfrac{\partial g}{\partial y}\right) \\\\ = \dfrac{\partial}{\partial x}(xe^{xy} + 2y\sin(x)) \\\\= e^{xy} + xye^{xy} + 2y\cos(x) \] The equality \(\dfrac{\partial^2 g}{\partial y \partial x} = \dfrac{\partial^2 g}{\partial x \partial y}\) illustrates Clairaut's theorem on the equality of mixed partials under certain continuity conditions.

a. With respect to \(x\): \[ \dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x}(x^3y^2) + \dfrac{\partial}{\partial x}(x^2e^y) \\\\= 3x^2y^2 + 2xe^y \] b. With respect to \(y\) \[ \dfrac{\partial f}{\partial y} = \dfrac{\partial}{\partial y}(x^3y^2) + \dfrac{\partial}{\partial y}(x^2e^y) \\\\= 2x^3y + x^2e^y \] 2. Second-Order Partial Derivatives

a. With respect to \(x\) twice: \[ \dfrac{\partial^2 f}{\partial x^2} = \dfrac{\partial}{\partial x} \left(\dfrac{\partial f}{\partial x}\right) \\\\ = \dfrac{\partial}{\partial x}(3x^2y^2 + 2xe^y) \\\\= 6xy^2 + 2e^y \] b. With respect to \(y\) twice: \[ \dfrac{\partial^2 f}{\partial y^2} = \dfrac{\partial}{\partial y} \left(\dfrac{\partial f}{\partial y}\right) \\\\ = \dfrac{\partial}{\partial y}(2x^3y + x^2e^y) \\\\= 2x^3 + x^2e^y \] c. With respect to \(x\) then \(y\): \[ \dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right) \\\\ = \dfrac{\partial}{\partial y}(3x^2y^2 + 2xe^y) \\\\ = 6x^2y + 2xe^y \] d. With respect to \(y\) then \(x\): \[ \dfrac{\partial^2 f}{\partial x \partial y} = \dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial y}\right)\\\\ = \dfrac{\partial}{\partial x}(2x^3y + x^2e^y) \\\\= 6x^2y + 2xe^y \] In this example we also have the equality \(\dfrac{\partial^2 f}{\partial y \partial x} = \dfrac{\partial^2 f}{\partial x \partial y}\), which illustrates Clairaut's theorem on the equality of mixed partials under certain continuity conditions.

3. Third-Order Partial Derivatives

a. With respect to \(x\) twice, then \(y\): \[ \dfrac{\partial^3 f}{\partial y \partial x^2} = \dfrac{\partial}{\partial y} \left(\dfrac{\partial^2 f}{\partial x^2} \right) \\\\ = \dfrac{\partial}{\partial y}(6xy^2 + 2e^y) = 12 x y + 2 e^y \] b. With respect to \(x\), then \(y\), then \(x\) again : \[ \dfrac{\partial^3 f}{\partial x \partial y \partial x} = \dfrac{\partial}{\partial x} \left(\dfrac{\partial^2 f}{\partial x \partial y} \right) \\\\ = \dfrac{\partial}{\partial x}(6x^2y + 2xe^y) = 12xy + 2e^y \]

Consider the function \( f(x, y) = x^2y + y^3 \).

First, let usfind the mixed partial derivatives:

1. Find \( \dfrac{{\partial f}}{{\partial x}} \): \[ \dfrac{{\partial f}}{{\partial x}} = 2xy \] 2. Find \( \dfrac{{\partial f}}{{\partial y}} \): \[ \dfrac{{\partial f}}{{\partial y}} = x^2 + 3y^2 \] Now, let's find the mixed partial derivatives:

1. \( \dfrac{{\partial^2 f}}{{\partial x \partial y}} \): \[ \dfrac{{\partial^2 f}}{{\partial x \partial y}} = \dfrac{{\partial}}{{\partial x}} \left( \dfrac{{\partial f}}{{\partial y}} \right) = \dfrac{{\partial}}{{\partial x}} (x^2 + 3y^2) = 2x \] 2. \( \dfrac{{\partial^2 f}}{{\partial y \partial x}} \): \[ \dfrac{{\partial^2 f}}{{\partial y \partial x}} = \dfrac{{\partial}}{{\partial y}} \left( \dfrac{{\partial f}}{{\partial x}} \right) = \dfrac{{\partial}}{{\partial y}} (2xy) = 2x \] As per Clairaut's theorem, since both mixed partial derivatives are continuous and equal, we have \( \dfrac{{\partial^2 f}}{{\partial x \partial y}} = \dfrac{{\partial^2 f}}{{\partial y \partial x}} \).

Functions of Many Variables