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Section 13.4 : Higher Order Partial Derivatives
7. Given \(f\left( {x,y,z} \right) = {x^4}{y^3}{z^6}\) find \(\displaystyle \frac{{{\partial ^6}f}}{{\partial y\partial {z^2}\partial y\partial {x^2}}}\).
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Start SolutionThrough a natural extension of Clairaut’s theorem we know we can do these partial derivatives in any order we wish to. However, in this case there doesn’t seem to be any reason to do anything other than the order shown in the problem statement.
Here is the first derivative we need to take.
\[\frac{{\partial f}}{{\partial x}} = 4{x^3}{y^3}{z^6}\] Show Step 2The second derivative is,
\[\frac{{{\partial ^2}f}}{{\partial {x^2}}} = 12{x^2}{y^3}{z^6}\] Show Step 3The third derivative is,
\[\frac{{{\partial ^3}f}}{{\partial y\partial {x^2}}} = 36{x^2}{y^2}{z^6}\] Show Step 4The fourth derivative is,
\[\frac{{{\partial ^4}f}}{{\partial z\partial y\partial {x^2}}} = 216{x^2}{y^2}{z^5}\] Show Step 5The fifth derivative is,
\[\frac{{{\partial ^5}f}}{{\partial {z^2}\partial y\partial {x^2}}} = 1080{x^2}{y^2}{z^4}\] Show Step 6The sixth and final derivative we need for this problem is,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{{\partial ^6}f}}{{\partial y\partial {z^2}\partial y\partial {x^2}}} = 2160{x^2}y{z^4}}}\]