What is the derivative of (2t-1)(5t-5)^{-1} ?

The derivative of (2t-1)(5t-5)^{-1} is -1/(5(t-1)^2)

What is the first derivative of (2t-1)(5t-5)^{-1} ?

The first derivative of (2t-1)(5t-5)^{-1} is -1/(5(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

(2 t - 1) (5 t - 5)^{- 1}

- \frac{1}{5 ( t - 1 ) ^{2}}

Solution steps

\frac{d}{d t} ((2 t - 1) (5 t - 5)^{- 1})

Apply the Product Rule:

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

f = 2 t - 1, g = (5 t - 5)^{- 1}

= \frac{d}{d t} (2 t - 1) (5 t - 5)^{- 1} + \frac{d}{d t} ((5 t - 5)^{- 1}) (2 t - 1)

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

\frac{d}{d t} ((5 t - 5)^{- 1}) = - \frac{1}{5 ( t - 1 ) ^{2}}

= 2 (5 t - 5)^{- 1} + (- \frac{1}{5 ( t - 1 ) ^{2}}) (2 t - 1)

Simplify

2 (5 t - 5)^{- 1} + (- \frac{1}{5 ( t - 1 ) ^{2}}) (2 t - 1) : - \frac{1}{5 ( t - 1 ) ^{2}}

= - \frac{1}{5 ( t - 1 ) ^{2}}

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

y^{^{''}} - y^{^{'}} - 2 y = - 2 t + 8 t^{2}

\frac{4}{( s - 1 ) ^{3}}

y^{^{'}} = a - b y^{2}

\frac{\partial}{\partial x} (cos (4 x))

What is the derivative of (2t-1)(5t-5)^{-1} ?
The derivative of (2t-1)(5t-5)^{-1} is -1/(5(t-1)^2)
What is the first derivative of (2t-1)(5t-5)^{-1} ?
The first derivative of (2t-1)(5t-5)^{-1} is -1/(5(t-1)^2)