How to find derivative of function y=(3x-e^x)/(3x+e^x)?

We'll just have to apply the quotient rule to find out the
derivative of the given function:

(f/g)' = (f'*g -
f*g')/`g^(2)`

Let f(x) = 3x - `e^(x)` => f'(x) = 3 -
`e^(x)`

Let g(x) = 3x + `e^(x)` => g'(x) = 3 +
`e^(x)`

dy/dx = [(3-`e^(x)` )(3x+`e^(x)`)- (3x-`e^(x)`
)(3+`e^(x)` )]/`(3x+e^(x))^(2)`

dy/dx = (9x + 3`e^(x)` -
3x`e^(x)`- `e^(2x)`- 9x-3x`e^(x)`+ 3`e^(x)` + `e^(2x)` )/
`(3x+e^(x))^(2)`

We'll combine and eliminate like
terms:

dy/dx = (6`e^(x)` - 6x`e^(x)`
)/`(3x+e^(x))^(2)`

dy/dx = 6`e^(x)` (1 -
x)/`(3x+e^(x))^(2)`

The requested derivative
of the function is dy/dx = 6`e^(x)` (1-x)/`(3x+e^(x))^(2)`
.