e to the x Diff Problem

[Jason76]

\(\displaystyle f(x) = \dfrac{8 - xe^{x}}{x + e^{x}}\)

\(\displaystyle f'(x) = \dfrac{[x + e^{x}][(e^{x})(1) + (x)(e^{x})] - [8 -xe^{x}][1 + e^{x}]}{(x + e^{x})^{2}}\)

\(\displaystyle f'(x) = \dfrac{[x + e^{x}][(e^{x} + xe^{x}] - [8 -xe^{x}][1 + e^{x}]}{(x + e^{x})^{2}}\)

 

[stapel]

\(\displaystyle f(x) = \dfrac{8 - xe^{x}}{x + e^{x}}\)

\(\displaystyle f'(x) = \dfrac{[x + e^{x}][(e^{x})(1) + (x)(e^{x})] - [8 -xe^{x}][1 + e^{x}]}{(x + e^{x})^{2}}\)

How are you getting that the derivative of \(\displaystyle xe^x\) is just \(\displaystyle e^x\)? Please show your steps for applying the Product Rule to this. Thank you!

 

[Jason76]

\(\displaystyle f(x) = \dfrac{8 - xe^{x}}{x + e^{x}}\)

\(\displaystyle f'(x) = \dfrac{[x + e^{x}][(e^{x})(1) + (x)(e^{x})] - [8 -xe^{x}][1 + e^{x}]}{(x + e^{x})^{2}}\)

\(\displaystyle f'(x) = \dfrac{[x + e^{x}][(e^{x} + xe^{x}] - [8 -xe^{x}][1 + e^{x}]}{(x + e^{x})^{2}}\)

How do we foil the numerator out correctly?

\(\displaystyle [xe^{x} + x^{2}e^{x} + (e^{x})^{2} + x(e^{x})^{2}] - [8 - 8e^{x} - xe^{x} - x(e^{x})^{2}]\)

 

[Deleted member 4993]

\(\displaystyle f(x) = \dfrac{8 - xe^{x}}{x + e^{x}}\)

\(\displaystyle f'(x) = \dfrac{[x + e^{x}][(e^{x})(1) + (x)(e^{x})] - [8 -xe^{x}][1 + e^{x}]}{(x + e^{x})^{2}}\)

\(\displaystyle f'(x) = \dfrac{[x + e^{x}][(e^{x} + xe^{x}] - [8 -xe^{x}][1 + e^{x}]}{(x + e^{x})^{2}}\)

How do we foil the numerator out correctly?

\(\displaystyle [xe^{x} + x^{2}e^{x} + (e^{x})^{2} + x(e^{x})^{2}] - [8 - 8e^{x} - xe^{x} - x(e^{x})^{2}]\) ........... Looks good to me

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[srmichael]

One little hiccup in the expansion of the numerator. The second set of parentheses should be:

\(\displaystyle [8+8e^x-xe^x-x(e^x)^2]\)