What is the derivative of y=(2t)/(3+t^2) ?

The derivative of y=(2t)/(3+t^2) is (2(3-t^2))/((3+t^2)^2)

What is the first derivative of y=(2t)/(3+t^2) ?

The first derivative of y=(2t)/(3+t^2) is (2(3-t^2))/((3+t^2)^2)

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derivative of

y = \frac{2 t}{3 + t ^{2}}

\frac{2 ( 3 - t ^{2} )}{( 3 + t ^{2} ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{2 t}{3 + t ^{2}})

Take the constant out:

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

= 2 \frac{d}{d t} (\frac{t}{3 + 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} ( 3 + t ^{2} ) - \frac{d}{d t} ( 3 + t ^{2} ) t}{( 3 + t ^{2} ) ^{2}}

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

\frac{d}{d t} (3 + t^{2}) = 2 t

= 2 \cdot \frac{1 \cdot ( 3 + t ^{2} ) - 2 tt}{( 3 + t ^{2} ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot ( 3 + t ^{2} ) - 2 tt}{( 3 + t ^{2} ) ^{2}} : \frac{2 ( 3 - t ^{2} )}{( 3 + t ^{2} ) ^{2}}

= \frac{2 ( 3 - t ^{2} )}{( 3 + t ^{2} ) ^{2}}

x \to - 1 lim (x^{2} + 1)

\frac{\partial}{\partial y} (e^{x y} x)

\frac{( - 10 x - 9 y x ^{2} )}{3 ( x ^{3} - 5 y ^{2} )}

\int x \cdot cos (2 x) d x

\int \frac{ln ( v )}{v ^{2} + 2 v + 1} d v

What is the derivative of y=(2t)/(3+t^2) ?
The derivative of y=(2t)/(3+t^2) is (2(3-t^2))/((3+t^2)^2)
What is the first derivative of y=(2t)/(3+t^2) ?
The first derivative of y=(2t)/(3+t^2) is (2(3-t^2))/((3+t^2)^2)