What is the derivative of (7-xe^x)/(x+e^x) ?

The derivative of (7-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-7e^x-7)/((x+e^x)^2)

What is the first derivative of (7-xe^x)/(x+e^x) ?

The first derivative of (7-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-7e^x-7)/((x+e^x)^2)

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derivative of

\frac{7 - x e ^{x}}{x + e ^{x}}

\frac{- e ^{x} x ^{2} - e ^{2 x} - 7 e ^{x} - 7}{( x + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{7 - 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} ( 7 - x e ^{x} ) ( x + e ^{x} ) - \frac{d}{d x} ( x + e ^{x} ) ( 7 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

\frac{d}{d x} (7 - 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} ) ( 7 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

Expand

(- e^{x} x - e^{x}) (x + e^{x}) - (1 + e^{x}) (7 - x e^{x}) : - e^{x} x^{2} - e^{2 x} - 7 e^{x} - 7

= \frac{- e ^{x} x ^{2} - e ^{2 x} - 7 e ^{x} - 7}{( x + e ^{x} ) ^{2}}

x \to 1 lim (\frac{x - 1}{x - 1})

\int (x y^{2} + x^{2} + y + 4) d x

(\frac{x}{x ^{5} + 1})^{^{'}}

f (v) = (v^{- 3} + 3 v^{- 2})^{3}

f^{^{'}} (x) = 7 x^{4}

What is the derivative of (7-xe^x)/(x+e^x) ?
The derivative of (7-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-7e^x-7)/((x+e^x)^2)
What is the first derivative of (7-xe^x)/(x+e^x) ?
The first derivative of (7-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-7e^x-7)/((x+e^x)^2)