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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

arccot (3 x)

n = 1 \sum \infty \frac{n !}{1 4 ^{n}}

\frac{\partial}{\partial y} (- 2 sec^{2} (3 x - 2 y + 1))

\frac{d}{d x} (tan (5 x + 1))

x y^{2} y^{^{'}} = x + 6

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