Gas of independent molecules
This is more of a maths question, but I can give a quick answer.

As you said,

$$\frac{\mathrm{d}}{\mathrm{d}Q}\ln Q = \frac{1}{Q}$$

You could write this as

$$\frac{\mathrm{d}(\ln Q)}{\mathrm{d}Q} = \frac{1}{Q}$$

So, by the chain rule,

$$\begin{align}
\frac{\partial (\ln Q)}{\partial V} &amp;= \frac{\partial Q}{\partial V}\cdot\frac{\mathrm{d}(\ln Q)}{\mathrm{d}Q} \\
&amp;= \frac{1}{Q}\frac{\partial Q}{\partial V}
\end{align}$$

If you are studying physical chemistry, it is a good idea to remember this trick. It appears everywhere.