What is the derivative of (3-xe^x)/(x+e^x) ?

The derivative of (3-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-3e^x-3)/((x+e^x)^2)

What is the first derivative of (3-xe^x)/(x+e^x) ?

The first derivative of (3-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-3e^x-3)/((x+e^x)^2)

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derivative of

\frac{3 - x e ^{x}}{x + e ^{x}}

\frac{- e ^{x} x ^{2} - e ^{2 x} - 3 e ^{x} - 3}{( x + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{3 - 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} ( 3 - x e ^{x} ) ( x + e ^{x} ) - \frac{d}{d x} ( x + e ^{x} ) ( 3 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

\frac{d}{d x} (3 - 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} ) ( 3 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

Expand

(- e^{x} x - e^{x}) (x + e^{x}) - (1 + e^{x}) (3 - x e^{x}) : - e^{x} x^{2} - e^{2 x} - 3 e^{x} - 3

= \frac{- e ^{x} x ^{2} - e ^{2 x} - 3 e ^{x} - 3}{( x + e ^{x} ) ^{2}}

\frac{d}{d x} (x cos (ln (x)))

\int 2 cos (θ) d θ

5 e^{x} cos (x)

\frac{d}{d x} (- sin (\frac{2}{x + 1} - e^{x^{2}} \cdot 2 x))

{\frac{3 !}{s ^{4} ( s + 2 ) ^{2}}}

What is the derivative of (3-xe^x)/(x+e^x) ?
The derivative of (3-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-3e^x-3)/((x+e^x)^2)
What is the first derivative of (3-xe^x)/(x+e^x) ?
The first derivative of (3-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-3e^x-3)/((x+e^x)^2)