What is the derivative of t^2(t+1)^{-1} ?

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

What is the first derivative of t^2(t+1)^{-1} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

t^{2} (t + 1)^{- 1}

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t^{2}, g = (t + 1)^{- 1}

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

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

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

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

Simplify

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

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

3 x^{\frac{1}{3}}

y^{^{''}} + 2 y^{^{'}} - 35 y = 0

x \to 5 - lim (\frac{8}{x - 5})

y = 6^{- 4 x}

\frac{d}{d t} (e^{t} t)

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