Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 13.4 : Higher Order Partial Derivatives
2. Verify Clairaut’s Theorem for the following function.
\[A\left( {x,y} \right) = \cos \left( {\frac{x}{y}} \right) - {x^7}{y^4} + {y^{10}}\]Show All Steps Hide All Steps
Start SolutionFirst, we know we’ll need the two 1st order partial derivatives. Here they are,
\[{A_x} = - \frac{1}{y}\sin \left( {\frac{x}{y}} \right) - 7{x^6}{y^4}\hspace{0.5in}{A_y} = \frac{x}{{{y^2}}}\sin \left( {\frac{x}{y}} \right) - 4{x^7}{y^3} + 10{y^9}\] Show Step 2Now let’s compute each of the mixed second order partial derivatives.
\[\begin{align*}{A_{x\,y}} & = {\left( {{A_x}} \right)_y} = \frac{1}{{{y^2}}}\sin \left( {\frac{x}{y}} \right) + \frac{x}{{{y^3}}}\cos \left( {\frac{x}{y}} \right) - 28{x^6}{y^3}\\ {A_{y\,x}} & = {\left( {{A_y}} \right)_x} = \frac{1}{{{y^2}}}\sin \left( {\frac{x}{y}} \right) + \frac{x}{{{y^3}}}\cos \left( {\frac{x}{y}} \right) - 28{x^6}{y^3}\end{align*}\]Okay, we can see that \({A_{x\,y}} = {A_{y\,x}}\) and so Clairaut’s theorem has been verified for this function.