Copy differences between two dirs to a a new dir?

If I understand you correctly, you want to compare files in two directories and copy the files with different contents to a third directory. I don't see a way to do that with just one command (unless you write your own script or function). But one of the strong points of the bash shell is that you can chain commands, using the output of one command as the input of the next.

First, you can use diff as usual:

diff -q dir1/ dir2/

That will give you an output along the lines of

Files dir1/file2.txt and dir2/file2.txt differ
Files dir1/file3.txt and dir2/file3.txt differ

Now, you need to extract the file paths from that output. You could use a number of commands to do that, cut, grep, sed, awk or others. I'll use awk in this example and extract the files from dir2:

diff -q dir1 dir2 | awk -n '{ print $4 }'

The output of this pipeline would be something like

dir2/file2.txt
dir2/file3.txt

That output can now be fed to xargs as follows:

diff -q dir1 dir2 | awk -n '{ print $4 }' | xargs -d '\n' -I {} cp {} target-dir/

Please note that this method only works when the differing files exist in both directories. If there's a file that only exists in one of the two directories, diffs output will be different, which will make cutting the filename out fail. Also, the file names must not contain any spaces, because that would trip up the cutting as well.