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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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)` .

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