What is the derivative of (5-xe^x/(x+e^x)) ?

The derivative of (5-xe^x/(x+e^x)) is (-e^xx^2-e^{2x}-5e^x-5)/((x+e^x)^2)

What is the first derivative of (5-xe^x/(x+e^x)) ?

The first derivative of (5-xe^x/(x+e^x)) is (-e^xx^2-e^{2x}-5e^x-5)/((x+e^x)^2)

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\frac{d}{d x} (\frac{5 - x e ^{x}}{x + e ^{x}})

\frac{- e ^{x} x ^{2} - e ^{2 x} - 5 e ^{x} - 5}{( x + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{5 - 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} ( 5 - x e ^{x} ) ( x + e ^{x} ) - \frac{d}{d x} ( x + e ^{x} ) ( 5 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

\frac{d}{d x} (5 - 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} ) ( 5 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

Expand

(- e^{x} x - e^{x}) (x + e^{x}) - (1 + e^{x}) (5 - x e^{x}) : - e^{x} x^{2} - e^{2 x} - 5 e^{x} - 5

= \frac{- e ^{x} x ^{2} - e ^{2 x} - 5 e ^{x} - 5}{( x + e ^{x} ) ^{2}}

1 + e^{x}

f (x) = 5 x^{2}, at x = 2

\frac{d}{d x} (\frac{1 + sin ( x )}{x cos ( x )})

x \to 1 lim (3^{(\frac{1}{x})})

3 sin (3 t + 2)

What is the derivative of (5-xe^x/(x+e^x)) ?
The derivative of (5-xe^x/(x+e^x)) is (-e^xx^2-e^{2x}-5e^x-5)/((x+e^x)^2)
What is the first derivative of (5-xe^x/(x+e^x)) ?
The first derivative of (5-xe^x/(x+e^x)) is (-e^xx^2-e^{2x}-5e^x-5)/((x+e^x)^2)