How do you use the quotient rule to find the derivative of $y = \frac{1 - x \cdot e^{x}}{x + e^{x}}$ $y=(1-x*e^x)/(x+e^x)$ ?

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Answer 1 · 2014-09-24T02:50:02

$y = \frac{f (x)}{g (x)} = \frac{1 - x e^{x}}{x + e^{x}}$ $y=f(x)/g(x)=(1-xe^x)/(x+e^x)$

$f'(x)=-(xe^x+e^x)$

$g'(x)=1+e^x$

$y'=(g(x)f'(x)-g'(x)f(x))/((g(x))^2)$

Substitute in the values for $f(x), f'(x), g(x), and g'(x)$

$=((x+e^x)(-xe^x-e^x)-(1+e^x)(1-xe^x))/(x+e^x)^2$

FOIL

$=(-x^2e^x-xe^x-xe^(2x)-e^(2x)-[1-xe^x+e^x-xe^(2x)])/(x+e^x)^2$

Distribute the negative

$=(-x^2e^x-xe^x-xe^(2x)-e^(2x)-1+xe^x-e^x+xe^(2x))/(x+e^x)^2$

Combine like terms

$=(-x^2e^x-e^x-e^(2x)-1)/(x+e^x)^2$

How do you use the quotient rule to find the derivative of y=(1-x*e^x)/(x+e^x)y=1−x⋅exx+exy=(1-x*e^x)/(x+e^x) ?