Equation of the tangent to the graph of function `f` at point `(x_0,f(x_(0) ) )` is

`y-f(x_0)=f'(x_0)(x-x_0)` **(1)**

So we first need to find derivation of `f`. For this we will need *chain rule* `(f(g(x)))'=f'(g(x))g'(x)`.

`f'(x)=cos(e^(3x^2-12)-1)e^(3x^2-12) cdot 6x+2`

`f'(2)=cos(e^(3cdot2^2-12)-1)e^(3cdot2^2-12)cdot6cdot2+2=cos(e^0-1)e^0cdot12+2=`

`cos0cdot12+2=14`

Since we already know that `f(2)=4` we can put all this into formula (1).

`y-4=14(x-2)`

`y=14x-28+4`

`y=14x-24`

**So equation of tangent line of graph of function `f` at point `(2,4)` is** `y=14x-24`**.**