What is the derivative of f(x)=(2x-1)(6x-6)^{-1} ?

The derivative of f(x)=(2x-1)(6x-6)^{-1} is -1/(6(x-1)^2)

What is the first derivative of f(x)=(2x-1)(6x-6)^{-1} ?

The first derivative of f(x)=(2x-1)(6x-6)^{-1} is -1/(6(x-1)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

f (x) = (2 x - 1) (6 x - 6)^{- 1}

- \frac{1}{6 ( x - 1 ) ^{2}}

Solution steps

\frac{d}{d x} ((2 x - 1) (6 x - 6)^{- 1})

Apply the Product Rule:

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

f = 2 x - 1, g = (6 x - 6)^{- 1}

= \frac{d}{d x} (2 x - 1) (6 x - 6)^{- 1} + \frac{d}{d x} ((6 x - 6)^{- 1}) (2 x - 1)

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

\frac{d}{d x} ((6 x - 6)^{- 1}) = - \frac{1}{6 ( x - 1 ) ^{2}}

= 2 (6 x - 6)^{- 1} + (- \frac{1}{6 ( x - 1 ) ^{2}}) (2 x - 1)

Simplify

2 (6 x - 6)^{- 1} + (- \frac{1}{6 ( x - 1 ) ^{2}}) (2 x - 1) : - \frac{1}{6 ( x - 1 ) ^{2}}

= - \frac{1}{6 ( x - 1 ) ^{2}}

f (x) = \frac{x ^{2}}{x + 2}

(tan (y)) d x - (cot (x)) d y = 0

\int \frac{x - 9}{( x + 5 ) ( x - 2 )} d x

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

2 e^{x - 2}

What is the derivative of f(x)=(2x-1)(6x-6)^{-1} ?
The derivative of f(x)=(2x-1)(6x-6)^{-1} is -1/(6(x-1)^2)
What is the first derivative of f(x)=(2x-1)(6x-6)^{-1} ?
The first derivative of f(x)=(2x-1)(6x-6)^{-1} is -1/(6(x-1)^2)