What is the d/(dt)((4+t)/(te^t)) ?

The d/(dt)((4+t)/(te^t)) is -(t^2+4t+4)/(e^tt^2)

What is the first d/(dt)((4+t)/(te^t)) ?

The first d/(dt)((4+t)/(te^t)) is -(t^2+4t+4)/(e^tt^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int x (6 x + 5)^{8} d x

f (x) = 2 e^{x} + 2 ln (x)

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

\int - \frac{1}{sin ( θ )} d θ

y^{^{'}} + t^{2} y = t^{2}

What is the d/(dt)((4+t)/(te^t)) ?
The d/(dt)((4+t)/(te^t)) is -(t^2+4t+4)/(e^tt^2)
What is the first d/(dt)((4+t)/(te^t)) ?
The first d/(dt)((4+t)/(te^t)) is -(t^2+4t+4)/(e^tt^2)