What is the derivative of (4+x/(xe^x)) ?

The derivative of (4+x/(xe^x)) is -(x^2+4x+4)/(e^xx^2)

What is the first derivative of (4+x/(xe^x)) ?

The first derivative of (4+x/(xe^x)) is -(x^2+4x+4)/(e^xx^2)

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\frac{d}{d x} (\frac{4 + x}{x e ^{x}})

- \frac{x ^{2} + 4 x + 4}{e ^{x} x ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 + 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} ( 4 + x ) x e ^{x} - \frac{d}{d x} ( x e ^{x} ) ( 4 + x )}{( x e ^{x} ) ^{2}}

\frac{d}{d x} (4 + x) = 1

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

= \frac{1 \cdot x e ^{x} - ( e ^{x} + e ^{x} x ) ( 4 + x )}{( x e ^{x} ) ^{2}}

Simplify

\frac{1 \cdot x e ^{x} - ( e ^{x} + e ^{x} x ) ( 4 + x )}{( x e ^{x} ) ^{2}} : - \frac{x ^{2} + 4 x + 4}{e ^{x} x ^{2}}

= - \frac{x ^{2} + 4 x + 4}{e ^{x} x ^{2}}

\int \frac{1}{3} e^{t} d t

f (x) = ln (\frac{5 x + 3}{5 x - 3})

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

\frac{\partial}{\partial y} (- 3 y)

y^{^{'}} + \frac{t}{t ^{2} - 16} y = \frac{e ^{t}}{t - 9}

What is the derivative of (4+x/(xe^x)) ?
The derivative of (4+x/(xe^x)) is -(x^2+4x+4)/(e^xx^2)
What is the first derivative of (4+x/(xe^x)) ?
The first derivative of (4+x/(xe^x)) is -(x^2+4x+4)/(e^xx^2)