In one line
Differentiating a partial derivative again gives second-order derivatives \(u_{xx},u_{yy}\) and the mixed ones \(u_{xy},u_{yx}\), which are equal whenever they are continuous.
Explained simply
Speed is the rate of change of distance, and acceleration is the rate of change of speed. In the same way, \(u_{xx}\) tells you how the slope in the \(x\) direction itself changes as you move in \(x\). The mixed derivative \(u_{xy}\) tells you how the \(x\)-slope changes as you move in \(y\). For nearly all functions you will meet, the order of differentiation does not matter: \(u_{xy}=u_{yx}\).
Technically, \(\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial x}\right)\), \(\dfrac{\partial^2u}{\partial x\,\partial y}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial y}\right)\). By Schwarz's (Clairaut's) theorem, if the mixed partial derivatives are continuous in a region, then \(\dfrac{\partial^2u}{\partial x\,\partial y}=\dfrac{\partial^2u}{\partial y\,\partial x}\) there. A function of two variables has four second-order derivatives, but only three are different in practice.
How it works
- Find \(u_x\) and \(u_y\).
- Differentiate \(u_x\) with respect to \(x\) to get \(u_{xx}\), and with respect to \(y\) to get \(u_{xy}\) (some books write \(u_{yx}\); the result is the same for smooth functions).
- Differentiate \(u_y\) with respect to \(y\) to get \(u_{yy}\).
- Use \(u_{xy}=u_{yx}\) as a check on your working.
- Many questions ask you to verify an equation such as Laplace's equation \(u_{xx}+u_{yy}=0\); compute each term and add.
Key formula
\(u_{xy}=u_{yx}\) (if continuous). Laplace's equation in two variables: \(\dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}=0\).
Quick example
Let \(u=x^3y^2+e^{xy}\).
- \(u_x=3x^2y^2+ye^{xy}\), \(\ u_y=2x^3y+xe^{xy}\)
- \(u_{xx}=6xy^2+y^2e^{xy}\), \(\ u_{yy}=2x^3+x^2e^{xy}\)
- From \(u_x\): \(\dfrac{\partial}{\partial y}u_x=6x^2y+e^{xy}+xye^{xy}\). From \(u_y\): \(\dfrac{\partial}{\partial x}u_y=6x^2y+e^{xy}+xye^{xy}\). Equal, as expected.
Another standard check: \(u=\log(x^2+y^2)\) gives \(u_{xx}=\dfrac{2(y^2-x^2)}{(x^2+y^2)^2}\), \(u_{yy}=\dfrac{2(x^2-y^2)}{(x^2+y^2)^2}\), so \(u_{xx}+u_{yy}=0\).
Common mistakes
- Forgetting the product rule when differentiating \(ye^{xy}\) with respect to \(y\).
- Thinking \(u_{xx}\) means \((u_x)^2\); it means differentiate twice.
- Not simplifying fractions before differentiating again, which makes the second derivative needlessly messy.
How MU asks it
Common 5/6-mark parts: "If \(u=\tan^{-1}\dfrac yx\), verify \(u_{xy}=u_{yx}\)", "Show that \(u=e^{x}\cos y\) satisfies Laplace's equation", or "If \(u=x^y\), prove \(u_{xxy}=u_{xyx}\)". For full marks show every first and second derivative separately, simplify each, and write the final comparison or substitution clearly.