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

`y=x(36^(-x))`

To take the derivative of y, apply product rule which is `(u*v)=v*u'+u*v'` .

So let,

`u=x` and `v=36^(-x)`

Then, take the derivative of u and v.

`u'=1`

To get v', apply the derivative of exponential functions which is `(a^u)=a^u*lna*u'` .

`v'=36^(-x)*ln36*(-x)'`

`v'=36^(-x)*ln36*-1`

`v'=-36^(-x)ln36`

And, plug-in u , v, u' and v' to the formula of product rule.

`y'=36^(-x)*1+ (-36^(-x)ln36)`

`y'=36^(-x)-36^(-x)ln36`

Express 36 with positive exponent.

`y'=1/36^x-(ln36)/36^x`

**Hence, the derivative of the given function is `y'=1/36^x-(ln36)/36^x` .**