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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 Clairautâ€™s theorem we know we can do these partial derivatives in any order we wish to. However, in this case there doesnâ€™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,

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