Let A = x i + y j - z k and B = 3x i + 2y j + z k. Compute the mixed second partial derivative of A x B with respect to y and then x, evaluated at the point (1, 2, 1).

Step-by-step solution

1. 1

   Do the cross product first, differentiate second. Differentiating a cross product component by component only works after you have the components, so evaluate the determinant

   $\vec A\times\vec B=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\ x&y&-z\\ 3x&2y&z\end{vmatrix}$

   The alternative — using the product rule $\partial_x(\vec A\times\vec B)=\partial_x\vec A\times\vec B+\vec A\times\partial_x\vec B$ — is valid but longer here.

2. 2

   Expand the determinant one component at a time.

   $\mathbf i:\ (y)(z)-(-z)(2y)=yz+2yz=3yz$
   $\mathbf j:\ -\big[(x)(z)-(-z)(3x)\big]=-\big[xz+3xz\big]=-4xz$
   $\mathbf k:\ (x)(2y)-(y)(3x)=2xy-3xy=-xy$

   so

   $\vec A\times\vec B=3yz\,\mathbf i-4xz\,\mathbf j-xy\,\mathbf k$

   The sign flip in front of the $\mathbf j$ term is part of the cofactor expansion and is the most common slip.

3. 3

   Differentiate with respect to $x$, treating $y$ and $z$ as constants.

   $\frac{\partial}{\partial x}\left(3yz\,\mathbf i-4xz\,\mathbf j-xy\,\mathbf k\right)=0\,\mathbf i-4z\,\mathbf j-y\,\mathbf k$

   The $\mathbf i$ component contains no $x$ at all, so it dies immediately.

4. 4

   Differentiate the result with respect to $y$.

   $\frac{\partial}{\partial y}\left(-4z\,\mathbf j-y\,\mathbf k\right)=0\,\mathbf j-1\,\mathbf k=-\mathbf k$

   Every surviving term was linear, so the second derivative is a constant vector: the answer is $-\mathbf k$ at every point, including $(1,2,1)$. The coordinates of the point were never needed — worth saying out loud in your answer.

5. 5

   Check with Clairaut's theorem by reversing the order. Differentiating first in $y$ gives $3z\,\mathbf i-x\,\mathbf k$, and then in $x$ gives $-\mathbf k$. The two orders agree, as they must for a polynomial (hence smooth) vector field.