What is the derivative of (1-xe^x/(1+e^x)) ?

The derivative of (1-xe^x/(1+e^x)) is (-e^xx-2e^x-e^{2x})/((1+e^x)^2)

What is the first derivative of (1-xe^x/(1+e^x)) ?

The first derivative of (1-xe^x/(1+e^x)) is (-e^xx-2e^x-e^{2x})/((1+e^x)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

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

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

\frac{\partial}{\partial x} (x^{2} y^{2} - y^{3} + 3 x^{4} + 5)

\int_{0}^{4} 10 x d x

f (x) = \frac{1}{x ^{2}}

y^{^{''}} + 2 k^{2} y^{^{'}} + k^{4} y = 0

\frac{d}{d x} (\frac{2 x sin ( x )}{2 + cos ( x )})

What is the derivative of (1-xe^x/(1+e^x)) ?
The derivative of (1-xe^x/(1+e^x)) is (-e^xx-2e^x-e^{2x})/((1+e^x)^2)
What is the first derivative of (1-xe^x/(1+e^x)) ?
The first derivative of (1-xe^x/(1+e^x)) is (-e^xx-2e^x-e^{2x})/((1+e^x)^2)