What is the d/(dt)((3t)/(1+t^3)) ?

The d/(dt)((3t)/(1+t^3)) is (3(1-2t^3))/((1+t^3)^2)

What is the first d/(dt)((3t)/(1+t^3)) ?

The first d/(dt)((3t)/(1+t^3)) is (3(1-2t^3))/((1+t^3)^2)

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\frac{d}{d t} (\frac{3 t}{1 + t ^{3}})

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

Solution steps

\frac{d}{d t} (\frac{3 t}{1 + t ^{3}})

Take the constant out:

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

= 3 \frac{d}{d t} (\frac{t}{1 + t ^{3}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{d}{d t} (1 + t^{3}) = 3 t^{2}

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

Simplify

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

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

x \to 0 + lim (ln (4 sin (x)))

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

\int \frac{1}{5 - 2 x} d x

t \to 0 lim (sin (7 t))

ln (9 x^{2} + 5 y^{2})

What is the d/(dt)((3t)/(1+t^3)) ?
The d/(dt)((3t)/(1+t^3)) is (3(1-2t^3))/((1+t^3)^2)
What is the first d/(dt)((3t)/(1+t^3)) ?
The first d/(dt)((3t)/(1+t^3)) is (3(1-2t^3))/((1+t^3)^2)