What is the derivative of y=(8-xe^x)/(x+e^x) ?

The derivative of y=(8-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-8e^x-8)/((x+e^x)^2)

What is the first derivative of y=(8-xe^x)/(x+e^x) ?

The first derivative of y=(8-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-8e^x-8)/((x+e^x)^2)

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derivative of

y = \frac{8 - x e ^{x}}{x + e ^{x}}

\frac{- e ^{x} x ^{2} - e ^{2 x} - 8 e ^{x} - 8}{( x + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{8 - 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} ( 8 - x e ^{x} ) ( x + e ^{x} ) - \frac{d}{d x} ( x + e ^{x} ) ( 8 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

\frac{d}{d x} (8 - 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} ) ( 8 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

Expand

(- e^{x} x - e^{x}) (x + e^{x}) - (1 + e^{x}) (8 - x e^{x}) : - e^{x} x^{2} - e^{2 x} - 8 e^{x} - 8

= \frac{- e ^{x} x ^{2} - e ^{2 x} - 8 e ^{x} - 8}{( x + e ^{x} ) ^{2}}

\frac{\partial}{\partial x} (\frac{2 F ( V , x , P )}{V ^{2} x P})

\frac{d ^{3}}{d x ^{3}} (\frac{5}{x})

y = (3 x^{6} - 2 x^{4})^{4}

\frac{\partial}{\partial y} (e^{x y z^{9}} y z^{9})

\int 2 e^{4} d x

What is the derivative of y=(8-xe^x)/(x+e^x) ?
The derivative of y=(8-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-8e^x-8)/((x+e^x)^2)
What is the first derivative of y=(8-xe^x)/(x+e^x) ?
The first derivative of y=(8-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-8e^x-8)/((x+e^x)^2)