Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 42

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)