e to the x quotient problem

[Jason76]

\(\displaystyle \dfrac{f(x)}{g(x)}\)

\(\displaystyle \dfrac{[g(x)][f'(x)] - [f(x)][g'(x)]}{g^{2}}\)

\(\displaystyle f(x) = \dfrac{e^{x}}{1 + x}\)

\(\displaystyle f'(x) = \dfrac{(1 + x)(\dfrac{d}{dx} e^{x}) - (e^{x})[\dfrac{d}{dx}(1 + x)]}{(1 + x)^{2}}\)

\(\displaystyle f'(x) = \dfrac{(1 + x)(e^{x})- (e^{x})(1)}{(1 + x)^{2}}\)

\(\displaystyle f'(x) = \dfrac{e^{x} + xe^{x} - e^{x}}{(1 + x)^{2}}\)

\(\displaystyle f'(x) = \dfrac{xe^{x}}{(1 + x)^{2}}\) - Is this the final answer

 

[Deleted member 4993]

\(\displaystyle \dfrac{f(x)}{g(x)}\)

\(\displaystyle \dfrac{[g(x)][f'(x)] - [f(x)][g'(x)]}{g^{2}}\)

\(\displaystyle f(x) = \dfrac{e^{x}}{1 + x}\)

\(\displaystyle f'(x) = \dfrac{(1 + x)(\dfrac{d}{dx} e^{x}) - (e^{x})[\dfrac{d}{dx}(1 + x)]}{(1 + x)^{2}}\)

\(\displaystyle f'(x) = \dfrac{(1 + x)(e^{x})- (e^{x})(1)}{(1 + x)^{2}}\)

\(\displaystyle f'(x) = \dfrac{e^{x} + xe^{x} - e^{x}}{(1 + x)^{2}}\)

\(\displaystyle f'(x) = \dfrac{xe^{x}}{(1 + x)^{2}}\) - Is this the final answer

Yes

 

[lookagain]

\(\displaystyle \dfrac{f(x)}{g(x)}\)

\(\displaystyle \dfrac{[g(x)][f'(x)] - [f(x)][g'(x)]}{g^{2}} \ \ \ \ \ \ \)

Do not type \(\displaystyle \ "g^2" \ \) for the denominator.

Be consistent with the forms and type the denominator as \(\displaystyle "[g(x)]^2," \ \ \) for instance.