What is the derivative of y=(x+x^2)/(e^x+1) ?

The derivative of y=(x+x^2)/(e^x+1) is (-e^xx^2+e^xx+2x+e^x+1)/((e^x+1)^2)

What is the first derivative of y=(x+x^2)/(e^x+1) ?

The first derivative of y=(x+x^2)/(e^x+1) is (-e^xx^2+e^xx+2x+e^x+1)/((e^x+1)^2)

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derivative of

y = \frac{x + x ^{2}}{e ^{x} + 1}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

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

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

\frac{d}{d x} (\frac{54}{x ^{4}})

\int 2 x + 4 x^{2} - \frac{5}{2} d x

\int \frac{x ^{2}}{2 ( 2 - x )} d x

x \to 0 lim (\frac{- 1}{14 x})

y = 6 x^{2} sin (x) tan (x)

What is the derivative of y=(x+x^2)/(e^x+1) ?
The derivative of y=(x+x^2)/(e^x+1) is (-e^xx^2+e^xx+2x+e^x+1)/((e^x+1)^2)
What is the first derivative of y=(x+x^2)/(e^x+1) ?
The first derivative of y=(x+x^2)/(e^x+1) is (-e^xx^2+e^xx+2x+e^x+1)/((e^x+1)^2)