What is the derivative of f(x)=(9-xe^x)/(x+e^x) ?

The derivative of f(x)=(9-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-9e^x-9)/((x+e^x)^2)

What is the first derivative of f(x)=(9-xe^x)/(x+e^x) ?

The first derivative of f(x)=(9-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-9e^x-9)/((x+e^x)^2)

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derivative of

f (x) = \frac{9 - x e ^{x}}{x + e ^{x}}

\frac{- e ^{x} x ^{2} - e ^{2 x} - 9 e ^{x} - 9}{( x + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{9 - 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} ( 9 - x e ^{x} ) ( x + e ^{x} ) - \frac{d}{d x} ( x + e ^{x} ) ( 9 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

\frac{d}{d x} (9 - 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} ) ( 9 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

Expand

(- e^{x} x - e^{x}) (x + e^{x}) - (1 + e^{x}) (9 - x e^{x}) : - e^{x} x^{2} - e^{2 x} - 9 e^{x} - 9

= \frac{- e ^{x} x ^{2} - e ^{2 x} - 9 e ^{x} - 9}{( x + e ^{x} ) ^{2}}

\frac{d}{d x} ((5 x - 2)^{3})

\frac{d}{d x} (sin^{2} (π - x))

\int \frac{a x}{x ^{2} - b x} d x

\frac{d y}{d t} = \frac{3 t ^{2} + t e ^{- t} + 1}{2 y - 4}

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

What is the derivative of f(x)=(9-xe^x)/(x+e^x) ?
The derivative of f(x)=(9-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-9e^x-9)/((x+e^x)^2)
What is the first derivative of f(x)=(9-xe^x)/(x+e^x) ?
The first derivative of f(x)=(9-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-9e^x-9)/((x+e^x)^2)