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Section 13.4 : Higher Order Partial Derivatives
1. Verify Clairaut’s Theorem for the following function.
\[f\left( {x,y} \right) = {x^3}{y^2} - \frac{{4{y^6}}}{{{x^3}}}\]Show All Steps Hide All Steps
Start SolutionFirst, we know we’ll need the two 1st order partial derivatives. Here they are,
\[{f_x} = 3{x^2}{y^2} + 12{x^{ - 4}}{y^6}\hspace{0.75in}{f_y} = 2{x^3}y - 24{x^{ - 3}}{y^5}\] Show Step 2Now let’s compute each of the mixed second order partial derivatives.
\[{f_{x\,y}} = {\left( {{f_x}} \right)_y} = 6{x^2}y + 72{x^{ - 4}}{y^5}\hspace{0.5in}{f_{y\,x}} = {\left( {{f_y}} \right)_x} = 6{x^2}y + 72{x^{ - 4}}{y^5}\]Okay, we can see that \({f_{x\,y}} = {f_{y\,x}}\) and so Clairaut’s theorem has been verified for this function.