What is the derivative of (4+t)/(te^t) ?

The derivative of (4+t)/(te^t) is -(t^2+4t+4)/(e^tt^2)

What is the first derivative of (4+t)/(te^t) ?

The first derivative of (4+t)/(te^t) is -(t^2+4t+4)/(e^tt^2)

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derivative of

\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 \frac{1}{t ^{3} t ^{2} - 1} d t

\int_{0}^{1} x 1 + 8 x^{2} d x

\int (x^{2} - 4 x + 3) d x

\int sin^{4} (2 x) cos^{3} (2 x) d x

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

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