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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d t} (2 t^{2} + 1) = 4 t

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

Simplify

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

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

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