What is the derivative of y=(6-xe^x)/(x+e^x) ?

The derivative of y=(6-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-6e^x-6)/((x+e^x)^2)

What is the first derivative of y=(6-xe^x)/(x+e^x) ?

The first derivative of y=(6-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-6e^x-6)/((x+e^x)^2)

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derivative of

y = \frac{6 - x e ^{x}}{x + e ^{x}}

\frac{- e ^{x} x ^{2} - e ^{2 x} - 6 e ^{x} - 6}{( x + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 - 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} ( 6 - x e ^{x} ) ( x + e ^{x} ) - \frac{d}{d x} ( x + e ^{x} ) ( 6 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

\frac{d}{d x} (6 - 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} ) ( 6 - x e ^{x} )}{( x + e ^{x} ) ^{2}}

Expand

(- e^{x} x - e^{x}) (x + e^{x}) - (1 + e^{x}) (6 - x e^{x}) : - e^{x} x^{2} - e^{2 x} - 6 e^{x} - 6

= \frac{- e ^{x} x ^{2} - e ^{2 x} - 6 e ^{x} - 6}{( x + e ^{x} ) ^{2}}

2 \cdot e^{(5 \cdot t - 2)}

\int_{\frac{1}{7}}^{5} 10 x ln (7 x) d x

\frac{\partial}{\partial x} (sin (\frac{y}{x + y}))

\frac{d}{d x} (\frac{cot ( x )}{x ^{2}})

\frac{d y}{d x} + \frac{1}{x} y = 7 x^{2}

What is the derivative of y=(6-xe^x)/(x+e^x) ?
The derivative of y=(6-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-6e^x-6)/((x+e^x)^2)
What is the first derivative of y=(6-xe^x)/(x+e^x) ?
The first derivative of y=(6-xe^x)/(x+e^x) is (-e^xx^2-e^{2x}-6e^x-6)/((x+e^x)^2)