What is the d/(dt)((e^{-t})/(1+t^2)) ?

The d/(dt)((e^{-t})/(1+t^2)) is (-e^{-t}(1+t^2)-2e^{-t}t)/((1+t^2)^2)

What is the first d/(dt)((e^{-t})/(1+t^2)) ?

The first d/(dt)((e^{-t})/(1+t^2)) is (-e^{-t}(1+t^2)-2e^{-t}t)/((1+t^2)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

f (x) = 4 x^{2} + 5 x, at x = - 3

\int \frac{4}{- x ^{2} + 6 x + 5} d x

arcsec (x) \cdot arccos (x)

\int - y^{2} d y

What is the d/(dt)((e^{-t})/(1+t^2)) ?
The d/(dt)((e^{-t})/(1+t^2)) is (-e^{-t}(1+t^2)-2e^{-t}t)/((1+t^2)^2)
What is the first d/(dt)((e^{-t})/(1+t^2)) ?
The first d/(dt)((e^{-t})/(1+t^2)) is (-e^{-t}(1+t^2)-2e^{-t}t)/((1+t^2)^2)