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 \({G_{y\,y\,y\,x\,x\,x\,y}}\).
Show All Steps Hide All Steps
Start SolutionThrough 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.
Show Step 2The 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 3The third derivative is,
\[{G_{x\,x\,x}} = - 8{y^4}\cos \left( {2x} \right)\] Show Step 4The fourth derivative is,
\[{G_{x\,x\,x\,y}} = - 32{y^3}\cos \left( {2x} \right)\] Show Step 5The fifth derivative is,
\[{G_{x\,x\,x\,y\,y}} = - 96{y^2}\cos \left( {2x} \right)\] Show Step 6The sixth derivative is,
\[{G_{x\,x\,x\,y\,y\,y}} = - 192y\cos \left( {2x} \right)\] Show Step 7The seventh and final derivative we need for this problem is,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{{G_{y\,y\,y\,x\,x\,x\,y}} = {G_{x\,x\,x\,y\,y\,y\,y}} = - 192\cos \left( {2x} \right)}}\]