What is the derivative of (2t)/(t^2+1) ?

The derivative of (2t)/(t^2+1) is (2(-t^2+1))/((t^2+1)^2)

What is the first derivative of (2t)/(t^2+1) ?

The first derivative of (2t)/(t^2+1) is (2(-t^2+1))/((t^2+1)^2)

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derivative of

\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}}

\int ∣ x + 2 ∣ d x

\frac{d}{d x} ((5 x^{2} - 3) \cdot (x^{2} + x + 4))

\frac{d}{d x} (e^{x} (12 x^{2} - 24 x + 24))

\int \frac{4 x - 3}{2 x ^{2} - 3 x - 14} d x

y^{^{'''}} + 8 y^{^{''}} + 5 y^{^{'}} - 50 y = 0

What is the derivative of (2t)/(t^2+1) ?
The derivative of (2t)/(t^2+1) is (2(-t^2+1))/((t^2+1)^2)
What is the first derivative of (2t)/(t^2+1) ?
The first derivative of (2t)/(t^2+1) is (2(-t^2+1))/((t^2+1)^2)