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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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Through 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 2

The second derivative is,

\[\frac{{{\partial ^2}f}}{{\partial {x^2}}} = 12{x^2}{y^3}{z^6}\] Show Step 3

The third derivative is,

\[\frac{{{\partial ^3}f}}{{\partial y\partial {x^2}}} = 36{x^2}{y^2}{z^6}\] Show Step 4

The fourth derivative is,

\[\frac{{{\partial ^4}f}}{{\partial z\partial y\partial {x^2}}} = 216{x^2}{y^2}{z^5}\] Show Step 5

The fifth derivative is,

\[\frac{{{\partial ^5}f}}{{\partial {z^2}\partial y\partial {x^2}}} = 1080{x^2}{y^2}{z^4}\] Show Step 6

The sixth and final derivative we need for this problem is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{{\partial ^6}f}}{{\partial y\partial {z^2}\partial y\partial {x^2}}} = 2160{x^2}y{z^4}}}\]