This is a product.

Let u = cos 2x and `v=e^(x^2-1)`

Then `(du)/(dx)=-2\ sin\ 2x`,

and `(dv)/(dx)=e^(x^2-1)(2x)`

So

`{: ((dy)/(dx),=[cos\ 2x][(e^(x^2-1))(2x)]+[(e^(x^2-1))(-2\ sin\ 2x)]),(,=(e^(x^2-1))[cos\ 2x(2x)-2\ sin\ 2x]),(,=2e^(x^2-1)[x\ cos\ 2x-sin\ 2x]) :}`