Prove that $\nabla\times (a\vec{A})= a(\nabla \times \vec{A})+(\nabla a)\times A$

Question

$\vec{A}$ is a vector field and each of its component is a function of $x, y$, and $z$:

$\vec{A} = u\hat{i} + v\hat{j} + w\hat{k}$

$u = u(x,y,z)$

$v = v(x,y,z)$

$w = w(x,y,z)$

$a$ is a scalar function of $x, y$, and $z$.

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\begin{equation}%Problem 4 \begin{aligned} \nabla\times\left(a\vec{A}\right) &= a\left(\nabla\times\vec{A}\right) + \left(\nabla a\right)\times\vec{A}\\ \nabla\times\left(a\vec{A}\right) &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\%[0.5em] a A_x & a A_y & a A_z \end{vmatrix}\\ &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\%[0.5em] au & av & aw \end{vmatrix}\\ &= \left(\frac{\partial \left(aw\right)}{\partial y} - \frac{\partial \left(av\right)}{\partial z}\right) - \left(\frac{\partial \left(aw\right)}{\partial x} - \frac{\partial \left(au\right)}{\partial z}\right) + \left(\frac{\partial \left(av\right)}{\partial x} - \frac{\partial \left(au\right)}{\partial y}\right)\\ &= a\frac{\partial^2w}{\partial y} + w\frac{\partial^2a}{\partial y} - a\frac{\partial^2v}{\partial z } - v\frac{\partial^2a}{\partial z} - a\frac{\partial^2w}{\partial x} - w\frac{\partial^2a}{\partial x} + a\frac{\partial^2v}{\partial z} + v\frac{\partial^2a}{\partial z} + a\frac{\partial^2v}{\partial x} + v\frac{\partial^2a}{\partial x} - a\frac{\partial^2u}{\partial y} - u\frac{\partial^2a}{\partial y} \\ &= a\left[\left(\frac{\partial^2w}{\partial y} - \frac{\partial^2v}{\partial z}\right) - \left(\frac{\partial^2w}{\partial x} - \frac{\partial^2u}{\partial z}\right) + \left(\frac{\partial^2v}{\partial x} - \frac{\partial^2u}{\partial y}\right)\right] + \left[\left(w \frac{\partial a}{\partial y} - v\frac{\partial a}{\partial z}\right) - w\left(\frac{\partial a }{\partial x} + u\frac{\partial a}{\partial z}\right) + v\left(\frac{\partial a}{\partial x} - u\frac{\partial a}{\partial y}\right)\right]\\ &= a\left[\left(\frac{\partial^2w}{\partial y} - \frac{\partial^2v}{\partial z}\right) - \left(\frac{\partial^2w}{\partial x} - \frac{\partial^2u}{\partial z}\right) + \left(\frac{\partial^2v}{\partial x} - \frac{\partial^2u}{\partial y}\right)\right] + \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \frac{\partial a}{\partial x} & \frac{\partial a}{\partial y} & \frac{\partial a}{\partial z}\\%[0.5em] u & v & w \end{vmatrix}\\ &= a\left(\nabla\times\vec{A}\right) + \left(\left(\nabla a\right)\times\vec{A}\right)\\ &= a\left(\nabla\times\vec{A}\right) + \left(\nabla a\right)\times\vec{A}\\ \end{aligned} \end{equation}

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Is it correct what I've done?

Answer 1

No the original matrix should be $$ \nabla\times(a\mathbf{A}) = \left|\begin{matrix}i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\aA_x & aA_y & aA_z \end{matrix}\right| $$ then preceed notice that derivatives are not applied to the components of $\mathbf{A}$

Answer 2

Maybe trivial and not really what you were looking for, but (in Einstein notation) $$\nabla\times(a{\bf A})=\epsilon_{ijk}\partial_j (a A_k)=\epsilon_{ijk}a(\partial_j A_k)+\epsilon_{ijk} (\partial_j a)A_k=a\nabla\times{\bf A}+(\nabla a)\times {\bf A}$$ The only rule you need is derivative of a product. Summation commutes with differentiation so indices (over which the expressions are summed) have no effect on differentiation.