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]`

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