Higher Order Partial Derivatives

Consider a two-variable function $f(x,y)$. Since both of the 1st-order partial derivatives are also functions of $x$ and $y$ we could in turn differentiate each with respect to $x$ or $y$. Therefore, in the case of a function of two variables there will be a total of 4 possible 2nd-order derivatives:

$$\begin{array}{rl}(f_x{)}_x& =f_{xx}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right)=\frac{{\partial }^{2}f}{\partial x^2}\\ (f_x{)}_y& =f_{xy}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{{\partial }^{2}f}{\partial y\partial x}\\ (f_y{)}_x& =f_{yx}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{{\partial }^{2}f}{\partial x\partial y}\\ (f_y{)}_y& =f_{yy}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial y}\right)=\frac{{\partial }^{2}f}{\partial y^2}\end{array}$$

• The 2nd and 3rd one are often called mixed partial derivatives since we are taking derivatives with respect to more than 1 variable

2nd-order Partial Derivative Example

Find all the second order derivatives for $f(x,y)=\cos (2x)-x^2{e}^{5}y+3y^2$

We’ll first need the first order derivatives:

\begin{align} f_{x}(x,y) &= - 2\sin(2x) - 2xe^5y \ f_{y}(x,y) &= -5x^{2}e^{5y} + 6y \ \end{align}

The 2nd-order derivatives are then:

$$\begin{array}{rl}{f}_{xx}& =-4\cos (2x)-2e^{5y}\\ f_{xy}& =-10x{e}^{5y}\\ f_{yx}& =-10x{e}^{5y}\\ f_{yy}& =-25x^2{e}^{5y}+6\end{array}$$

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Clairaut’s Theorem

When functions are “nice enough” they will have $f_{xy}=f_{yx}$.

Clairaut's Theorem

Suppose that $f$ is defined on a disk $D$ that contains the point $(a,b)$. If the functions $f_{xy}$ and $f_{yx}$ are continuous on this disk then, $f_{xy}(a,b)=f_{yx}(a,b)$

This applies to most functions where the two mixed 2nd-order derivatives are continuous then they will be equal.

This also works for higher orders: $f_{xxy}=f_{xyx}=f_{yxx}$ And also for more variables: $f_{xz}(x,y,z)=f_{zx}(x,y,z)$

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Higher Orders

For example, third order:

$$\begin{array}{rl}{f}_{xyx}& =(f_{xy}{)}_x=\frac{\partial }{\partial x}\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)=\frac{{\partial }^{3}f}{\partial x\partial y\partial x}\\ f_{yxx}& =(f_{yx}{)}_x=\frac{\partial }{\partial x}\left(\frac{{\partial }^{2}f}{\partial y\partial x}\right)=\frac{{\partial }^{3}f}{\partial x^2\partial y}\end{array}$$