Section 13.4 : Higher Order Partial Derivatives
6. Find all 2nd order derivatives for the following function.
\[f\left( {x,y,z} \right) = \frac{{{x^2}{y^6}}}{{{z^3}}} - 2{x^6}z + 8{y^{ - 3}}{x^4} + 4{z^2}\]Show All Steps Hide All Steps
Start SolutionFirst, we know we’ll need the three 1st order partial derivatives. Here they are,
\[\begin{align*}{f_x} & = 2x{y^6}{z^{ - 3}} - 12{x^5}z + 32{y^{ - 3}}{x^3}\\ {f_y} & = 6{x^2}{y^5}{z^{ - 3}} - 24{y^{ - 4}}{x^4}\\ {f_z} & = - 3{x^2}{y^6}{z^{ - 4}} - 2{x^6} + 8z\end{align*}\] Show Step 2Now let’s compute each of the second order partial derivatives (and there will be a few of them….).
\[\begin{align*}{f_{x\,x}} & = {\left( {{f_x}} \right)_x} = 2{y^6}{z^{ - 3}} - 60{x^4}z + 96{y^{ - 3}}{x^2}\\ {f_{x\,y}} & = {\left( {{f_x}} \right)_y} = 12x{y^5}{z^{ - 3}} - 96{y^{ - 4}}{x^3}\\ {f_{x\,z}} & = {\left( {{f_x}} \right)_z} = - 6x{y^6}{z^{ - 4}} - 12{x^5}\\ {f_{y\,x}} & = {f_{x\,y}} = 12x{y^5}{z^{ - 3}} - 96{y^{ - 4}}{x^3}\hspace{0.5in}{\mbox{by Clairaut's Theorem}}\\ {f_{y\,y}} & = {\left( {{f_y}} \right)_y} = 30{x^2}{y^4}{z^{ - 3}} + 96{y^{ - 5}}{x^4}\\ {f_{y\,z}} & = {\left( {{f_y}} \right)_z} = - 18{x^2}{y^5}{z^{ - 4}}\\ {f_{z\,x}} & = {f_{x\,z}} = - 6x{y^6}{z^{ - 4}} - 12{x^5}\hspace{0.75in}{\mbox{by Clairaut's Theorem}}\\ {f_{z\,y}} & = {f_{y\,z}} = - 18{x^2}{y^5}{z^{ - 4}}\hspace{1.25in}{\mbox{by Clairaut's Theorem}}\\ {f_{z\,z}} & = {\left( {{f_z}} \right)_z} = 12{x^2}{y^6}{z^{ - 5}} + 8\end{align*}\]Note that when we used Clairaut’s Theorem here we used the natural extension to the Theorem we gave in the notes.