How to calculate $\nabla \ln(|\mathbf{u}-\mathbf{v}|)$ I need to calculate:$$\nabla \ln(|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|)$$where $\mathbf{u}$ and $\mathbf{v}$ are vector valued functions with $\alpha$ and $\gamma$ as independent values and $|\cdot|$ denotes Euclidean norm.
Does the $\nabla$ operator represents in this case, $$\nabla = \frac{\partial}{\partial \alpha}\mathbf{e}_1+\frac{\partial}{\partial \gamma}\mathbf{e}_2?$$
If so, are the following calculations correct? NOT CORRECT, SEE EDIT
$$\frac{\partial}{\partial\alpha}\ln(|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|) = \frac{1}{|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|}\frac{1}{2|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|}2(\mathbf{u}(\alpha)-\mathbf{v}(\gamma))=\frac{1}{\mathbf{u}(\alpha)-\mathbf{v}(\gamma)},$$
$$\frac{\partial}{\partial\gamma}\ln(|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|) = \frac{1}{|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|}\frac{-1}{2|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|}2(\mathbf{u}(\alpha)-\mathbf{v}(\gamma))=\frac{-1}{\mathbf{u}(\alpha)-\mathbf{v}(\gamma)},$$
Then,
$$\nabla \ln(|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|) \stackrel{?}{=} \left[\frac{1}{\mathbf{u}(\alpha)-\mathbf{v}(\gamma)},\frac{-1}{\mathbf{u}(\alpha)-\mathbf{v}(\gamma)}\right].$$
EDIT
I do not think is correct because $\frac{1}{|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|}\frac{1}{2|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|}$ is a scalar but $2(\mathbf{u}(\alpha)-\mathbf{v}(\gamma))$ is a vector. Here is my new attempt:
$$\frac{\partial}{\partial\alpha}\ln(|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|) = \frac{1}{|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|}\frac{1}{2|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|}\left(2(u_1(\alpha)-v_1(\gamma))\frac{\mathrm{d}}{\mathrm{d}\alpha}u_1(\alpha)+2(u_2(\alpha)-v_2(\gamma))\frac{\mathrm{d}}{\mathrm{d}\alpha}u_2(\alpha)\right)$$where $u_1$ denotes the first component of vector $\mathbf{u}(\alpha)$.
Thanks for helping!
 A: 
Does the $\nabla$ operator represent in this case,
  $\nabla = \frac{\partial}{\partial \alpha}\mathbf{e}_1+\frac{\partial}{\partial \gamma}\mathbf{e}_2$?

Yes, that is a natural interpretation. In general, symbols mean whatever the author of text wants. Some people use notation like $\nabla_{x}$ or $\nabla_{x,t}$ to disambiguate. If you want $\nabla$ to mean that expression above, then it means that. 
To find partial derivatives, it is advisable to rewrite the formula as
$$\frac12 \ln(|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|^2)
$$
because the squared norm is just the inner product $\sum_k (u_k(\alpha)-v_k(\gamma)^2)$. I'll differentiate with respect to $\alpha$: 
$$\frac{\partial}{\partial\alpha} \sum_k (u_k(\alpha)-v_k(\gamma))^2
= \sum_k 2u_k'(\alpha)(u_k(\alpha)-v_k(\gamma)) = 2 \mathbf u'(\alpha)\cdot (\mathbf{u}(\alpha)-\mathbf{v}(\gamma))$$
By the chain rule,
$$\frac{\partial}{\partial\alpha} 
\frac12 \ln(|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|^2)
= \frac{\mathbf u'(\alpha)\cdot (\mathbf{u}(\alpha)-\mathbf{v}(\gamma))}{|\mathbf{u}(\alpha)-\mathbf{v}(\gamma)|^{2}}$$
The derivative for $\gamma$ can be obtained just by juggling symbols: swap $\mathbf{u}$ and $\mathbf{v}$; also swap $\alpha$ and $\gamma$.
