The problem goes as follows: Show that, under a rotation $\vec {\nabla} = \hat{i} \frac{\partial}{\partial x}+ \hat{j} \frac{\partial}{\partial y}+ \hat{k} \frac{\partial}{\partial z} = \hat{i'} \frac{\partial}{\partial x'}+ \hat{j'} \frac{\partial}{\partial y'}+ \hat{k'} \frac{\partial}{\partial z'} = \vec{\nabla '} $
My approach: We know, for a pure rotation: $ \hat {i} = l_1 \hat{i'} + l_2 \hat{j'} + l_3 \hat {k'}$ and so on... where $l_i , m_i, n_i$ are the direction cosines of the old co-ordinate axes (in respective order) w.r.t the new ones. Plugging the values in, we get: $\vec {\nabla} = \hat{i} \frac{\partial}{\partial x}+ \hat{j} \frac{\partial}{\partial y}+ \hat{k} \frac{\partial}{\partial z} = \sum [l_1 \frac {\partial}{\partial x} +m_1 \frac {\partial}{\partial y} +n_1 \frac {\partial}{\partial z}] \hat{i'} = \sum \frac {\partial}{\partial x'} \hat{i'} $
Is my approach alright? Is the inference at the very end correct?