What is the derivative of (2t)/(4+t^2) ?

The derivative of (2t)/(4+t^2) is (2(4-t^2))/((4+t^2)^2)

What is the first derivative of (2t)/(4+t^2) ?

The first derivative of (2t)/(4+t^2) is (2(4-t^2))/((4+t^2)^2)

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derivative of

\frac{2 t}{4 + t ^{2}}

\frac{2 ( 4 - t ^{2} )}{( 4 + t ^{2} ) ^{2}}

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

x \to 0 lim (\frac{5 x ^{3}}{x ^{3}})

x y^{2} d y = (x^{3} + y^{3}) d x

\frac{d}{d x} ((2 x^{2} - 1)^{3})

n = 1 \sum \infty \frac{2 n}{5 n - 7}

\frac{d}{d x} (2^{x^{3} - 2} - \frac{1}{x ^{2}})

What is the derivative of (2t)/(4+t^2) ?
The derivative of (2t)/(4+t^2) is (2(4-t^2))/((4+t^2)^2)
What is the first derivative of (2t)/(4+t^2) ?
The first derivative of (2t)/(4+t^2) is (2(4-t^2))/((4+t^2)^2)