What is the derivative of (3x-1)/((x-1)^3) ?

The derivative of (3x-1)/((x-1)^3) is -(6x)/((x-1)^4)

What is the first derivative of (3x-1)/((x-1)^3) ?

The first derivative of (3x-1)/((x-1)^3) is -(6x)/((x-1)^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{3 x - 1}{( x - 1 ) ^{3}}

- \frac{6 x}{( x - 1 ) ^{4}}

Solution steps

\frac{d}{d x} (\frac{3 x - 1}{( x - 1 ) ^{3}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 x - 1 ) ( x - 1 ) ^{3} - \frac{d}{d x} ( ( x - 1 ) ^{3} ) ( 3 x - 1 )}{( ( x - 1 ) ^{3} ) ^{2}}

\frac{d}{d x} (3 x - 1) = 3

\frac{d}{d x} ((x - 1)^{3}) = 3 (x - 1)^{2}

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

Simplify

\frac{3 ( x - 1 ) ^{3} - 3 ( x - 1 ) ^{2} ( 3 x - 1 )}{( ( x - 1 ) ^{3} ) ^{2}} : - \frac{6 x}{( x - 1 ) ^{4}}

= - \frac{6 x}{( x - 1 ) ^{4}}

\int 6 x + 2 d x

(x + 2)^{3} - e^{3 x}

y^{^{''}} + 7 y^{^{'}} + 10 y = - 20 t e^{4 t}

f (x) = \frac{1}{1 + e ^{- x}}

\frac{d}{d t} (\frac{1}{2} g (t) t^{2})

What is the derivative of (3x-1)/((x-1)^3) ?
The derivative of (3x-1)/((x-1)^3) is -(6x)/((x-1)^4)
What is the first derivative of (3x-1)/((x-1)^3) ?
The first derivative of (3x-1)/((x-1)^3) is -(6x)/((x-1)^4)