What is the derivative of y=(4-xe^x)/(x+e^x) ?

The derivative of y=(4-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-4e^x-4)/((x+e^x)^2)

What is the first derivative of y=(4-xe^x)/(x+e^x) ?

The first derivative of y=(4-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-4e^x-4)/((x+e^x)^2)

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derivative of

y = \frac{4 - x e ^{x}}{x + e ^{x}}

\frac{- e ^{x} x ^{2} - e ^{2 x} - 4 e ^{x} - 4}{( x + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 - 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} ( 4 - x e ^{x} ) ( x + e ^{x} ) - \frac{d}{d x} ( x + e ^{x} ) ( 4 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

\frac{d}{d x} (4 - 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} ) ( 4 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

Expand

(- e^{x} x - e^{x}) (x + e^{x}) - (1 + e^{x}) (4 - x e^{x}) : - e^{x} x^{2} - e^{2 x} - 4 e^{x} - 4

= \frac{- e ^{x} x ^{2} - e ^{2 x} - 4 e ^{x} - 4}{( x + e ^{x} ) ^{2}}

\int x^{2} \frac{1}{x} d x

\int_{4}^{7} \frac{1}{x ^{2} - 1} d x

\frac{d y}{d x} = 6 x + 2

\frac{\partial}{\partial y} (x^{5} + x y^{8} + 6)

y^{^{'}} = x (y + 1)

What is the derivative of y=(4-xe^x)/(x+e^x) ?
The derivative of y=(4-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-4e^x-4)/((x+e^x)^2)
What is the first derivative of y=(4-xe^x)/(x+e^x) ?
The first derivative of y=(4-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-4e^x-4)/((x+e^x)^2)