Recall that the derivative of a function f at a point x is denoted as `y' = f'(x)` .

There basic properties and formula we can apply to simplify a function.

For the problem `y = x(6^(-2x)),` we may apply the Product Rule for derivative:

Product Rule provides the formula:

`y = h(x)g(x)` then the derivative: `y'= h'(x)*g(x) + h(x)*g'(x)` .

In the problem, `y = x(6^(-2x))` , we let:

`h(x)=x` and` g(x) = 6^(-2x)` .

Derivative of each function:

`h'(x)= 1`

For the other function `g(x)=6^(-2x)` , we apply derivative of exponential function that follows: `d/(du)(a^u) =a^u* ln(a)*du` where `a!=1`

Then,

`g'(x)=6^(-2x)*ln(6) *(-2 )` .

We now have:

`h(x) =x`

`h'(x) = 1 `

`g(x)= 6^(-2x)`

`g'(x)=6^(-2x)*ln(6) *(-2) or(-2)(6^(-2x)) ln(6)`

Then applying the Product Rule: `y' =h'(x) g(x)+ h(x)* g'(x)` , we get:

`y'=1*6^(-2x)+(-2)(6^(-2x)) ln(6) *x`

`y' = 6^(-2x) -(2)(6^(-2x)) xln(6)`

It can be express in another form.

We can let:

`6^(-2x) = (6^2)^(-x) = 36^(-x)`

`6^(-2x) (2)= (3*2)^(-2x)(2)`

` = 3^(-2x)*2^(-2x)*2`

`= (3^2)^(-x) *2^(-2x+1)`

` = 9^(-x)*2^(-2x+1)`

`y' = 6^(-2x) -2(6^(-2x)) xln(6) ` becomes:

`y' = 36^(-x) - 9^(-x)*2^(-2x+1)xln(6) `