What is the d/(dt)((2t)/(1+t^2)) ?

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

What is the first d/(dt)((2t)/(1+t^2)) ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d t}{d t} = 1

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

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

Simplify

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

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

\int_{- 3}^{3} \frac{6}{x ^{2} + 9} d x

\int x^{9} \cdot e^{x^{10} - 9} d x

y = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

n = 0 \sum \infty e^{- \frac{9 n}{2}}

\int u^{- 3} d u

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