What is the derivative of (2xe^x/(x^2+1)) ?

The derivative of (2xe^x/(x^2+1)) is (2(e^xx^3-e^xx^2+e^xx+e^x))/((x^2+1)^2)

What is the first derivative of (2xe^x/(x^2+1)) ?

The first derivative of (2xe^x/(x^2+1)) is (2(e^xx^3-e^xx^2+e^xx+e^x))/((x^2+1)^2)

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

\frac{2 ( e ^{x} x ^{3} - e ^{x} x ^{2} + e ^{x} x + e ^{x} )}{( x ^{2} + 1 ) ^{2}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

2 \cdot \frac{( e ^{x} + e ^{x} x ) ( x ^{2} + 1 ) - 2 xx e ^{x}}{( x ^{2} + 1 ) ^{2}} : \frac{2 ( e ^{x} x ^{3} - e ^{x} x ^{2} + e ^{x} x + e ^{x} )}{( x ^{2} + 1 ) ^{2}}

= \frac{2 ( e ^{x} x ^{3} - e ^{x} x ^{2} + e ^{x} x + e ^{x} )}{( x ^{2} + 1 ) ^{2}}

\int 7 x^{2} sin (2 x) d x

k = 1 \sum \infty (\frac{1}{2})^{k}

\frac{d}{d x} (\frac{x ^{2}}{64} + \frac{y ^{2}}{25}) = 1

x \to 2 - lim (\frac{- 1}{x ^{2} - 4})

\int \frac{14 x - 1}{( 4 x + 1 ) ( x - 2 )} d x

What is the derivative of (2xe^x/(x^2+1)) ?
The derivative of (2xe^x/(x^2+1)) is (2(e^xx^3-e^xx^2+e^xx+e^x))/((x^2+1)^2)
What is the first derivative of (2xe^x/(x^2+1)) ?
The first derivative of (2xe^x/(x^2+1)) is (2(e^xx^3-e^xx^2+e^xx+e^x))/((x^2+1)^2)