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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

e^{5 x} cos (8 x)

h \to 0 lim ((2 h)^{2})

\frac{d}{d x} (sin^{c o s (x)} (x))

y = 8 e^{x} cos^{2} (x), at x = \frac{π}{4}

\frac{\partial}{\partial x} (- 2 cos (- 3 x + 7))

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