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### Section 13.4 : Higher Order Partial Derivatives

9. Given $$G\left( {x,y} \right) = {y^4}\sin \left( {2x} \right) + {x^2}{\left( {{y^{10}} - \cos \left( {{y^2}} \right)} \right)^7}$$ find $${G_{y\,y\,y\,x\,x\,x\,y}}$$.

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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.

In this case the $$y$$ derivatives of the second term will become unpleasant at some point given that we have four of them. However, the second term has an $${x^2}$$and there are three $$x$$ derivatives we’ll need to do eventually. Therefore, the second term will differentiate to zero with the third $$x$$ derivative. So, let’s make heavy use of Clairaut’s to do the three $$x$$ derivatives first prior to any of the $$y$$ derivatives so we won’t need to deal with the “messy” $$y$$ derivatives with the second term.

Here is the first derivative we need to take.

${G_x} = 2{y^4}cos\left( {2x} \right) + 2x{\left( {{y^{10}} - \cos \left( {{y^2}} \right)} \right)^7}$

Note that if we’d done a couple of $$y$$ derivatives first the second would have been a product rule and because we did the $$x$$ derivative first we won’t need to every work about the “messy” $$u$$ derivatives of the second term.

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The second derivative is,

${G_{x\,x}} = - 4{y^4}sin\left( {2x} \right) + 2{\left( {{y^{10}} - \cos \left( {{y^2}} \right)} \right)^7}$ Show Step 3

The third derivative is,

${G_{x\,x\,x}} = - 8{y^4}\cos \left( {2x} \right)$ Show Step 4

The fourth derivative is,

${G_{x\,x\,x\,y}} = - 32{y^3}\cos \left( {2x} \right)$ Show Step 5

The fifth derivative is,

${G_{x\,x\,x\,y\,y}} = - 96{y^2}\cos \left( {2x} \right)$ Show Step 6

The sixth derivative is,

${G_{x\,x\,x\,y\,y\,y}} = - 192y\cos \left( {2x} \right)$ Show Step 7

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

$\require{bbox} \bbox[2pt,border:1px solid black]{{{G_{y\,y\,y\,x\,x\,x\,y}} = {G_{x\,x\,x\,y\,y\,y\,y}} = - 192\cos \left( {2x} \right)}}$