What is the derivative of f(x)=(2-xe^x)/(x+e^x) ?

The derivative of f(x)=(2-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-2e^x-2)/((x+e^x)^2)

What is the first derivative of f(x)=(2-xe^x)/(x+e^x) ?

The first derivative of f(x)=(2-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-2e^x-2)/((x+e^x)^2)

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derivative of

f (x) = \frac{2 - x e ^{x}}{x + e ^{x}}

\frac{- e ^{x} x ^{2} - e ^{2 x} - 2 e ^{x} - 2}{( x + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 - x e ^{x}}{x + e ^{x}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( 2 - x e ^{x} ) ( x + e ^{x} ) - \frac{d}{d x} ( x + e ^{x} ) ( 2 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

\frac{d}{d x} (2 - x e^{x}) = - e^{x} x - e^{x}

\frac{d}{d x} (x + e^{x}) = 1 + e^{x}

= \frac{( - e ^{x} x - e ^{x} ) ( x + e ^{x} ) - ( 1 + e ^{x} ) ( 2 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

Expand

(- e^{x} x - e^{x}) (x + e^{x}) - (1 + e^{x}) (2 - x e^{x}) : - e^{x} x^{2} - e^{2 x} - 2 e^{x} - 2

= \frac{- e ^{x} x ^{2} - e ^{2 x} - 2 e ^{x} - 2}{( x + e ^{x} ) ^{2}}

e^{x} \cdot cos (x)

\int_{- 1}^{1} (\frac{1}{1 + x ^{2}})^{2} d x

\int \frac{3 x ^{3} - 4 x ^{2} + 3 x}{x ^{2} + 1} d x

\int 6 x^{5} - 7 x^{4} - 9 x^{2} d x

\int x + c d x

What is the derivative of f(x)=(2-xe^x)/(x+e^x) ?
The derivative of f(x)=(2-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-2e^x-2)/((x+e^x)^2)
What is the first derivative of f(x)=(2-xe^x)/(x+e^x) ?
The first derivative of f(x)=(2-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-2e^x-2)/((x+e^x)^2)