Pulling dx out of cos(x+dx)

It doesn't look ok. Can you perhaps write the whole thing/problem ?

 

I am going to assume that your "dx" is not a differential but just some slight change in x. [itex]\Delta x[/itex] would be better notation.

The best you can do is use the sum identities:
sin(x+ dx)= sin(x)cos(dx)+ cos(x)sin(dx)
cos(x+ dx)= cos(x)cos(dx)- sin(x)sin(dx)

You could then argue that for dx sufficiently small sin(dx) is approximately equal to dx and cos(dx) is approximately equal to 1:
sin(x+ dx) is approximately sin(x)+ cos(x)dx
cos(x+ dx) is approximately cos(x)- sin(x)dx.

Now, r sin(x+ dx)cos(x+ dx) will be approximately
[itex]sin(x)cos(x)+ cos^2(x)dx- sin^2(dx)- sin(x)cos(x)dx^2[/itex]

I suppose you could not argue that, for dx a "differential", you can drop [itex]dx^2[/itex] to get [itex]sin(x)cos(x)+ cos^2(x)dx- sin^2(x)dx[/itex]

But those still will not give you something "times dx".


As dextercioby said, please show the entire problem.