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

The d/(dt)((3t^2)/(1+t^3)) is (3(2t-t^4))/((1+t^3)^2)

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

The first d/(dt)((3t^2)/(1+t^3)) is (3(2t-t^4))/((1+t^3)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\frac{3 ( 2 t - t ^{4} )}{( 1 + t ^{3} ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

3 \cdot \frac{2 t ( 1 + t ^{3} ) - 3 t ^{2} t ^{2}}{( 1 + t ^{3} ) ^{2}} : \frac{3 ( 2 t - t ^{4} )}{( 1 + t ^{3} ) ^{2}}

= \frac{3 ( 2 t - t ^{4} )}{( 1 + t ^{3} ) ^{2}}

\frac{d}{d x} (x^{3} + 6 x^{2} + 8 x - 96)

\int \frac{1}{( x - 8 ) ^{2}} d x

\frac{\partial}{\partial y} (\frac{1}{1 + e ^{- x - 2 y}})

\frac{d}{d x} (3 (5 - 8 x)^{5})

\frac{d}{d x} (6 x + 3)

What is the d/(dt)((3t^2)/(1+t^3)) ?
The d/(dt)((3t^2)/(1+t^3)) is (3(2t-t^4))/((1+t^3)^2)
What is the first d/(dt)((3t^2)/(1+t^3)) ?
The first d/(dt)((3t^2)/(1+t^3)) is (3(2t-t^4))/((1+t^3)^2)