What is the derivative of f(x)=(-3x+1)/(x^2-x) ?

The derivative of f(x)=(-3x+1)/(x^2-x) is (3x^2-2x+1)/((x^2-x)^2)

What is the first derivative of f(x)=(-3x+1)/(x^2-x) ?

The first derivative of f(x)=(-3x+1)/(x^2-x) is (3x^2-2x+1)/((x^2-x)^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) = \frac{- 3 x + 1}{x ^{2} - x}

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

Solution steps

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

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

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

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

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

Simplify

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

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

y = \frac{ln ( x )}{x ^{5}}

\int \frac{e ^{3 t}}{t} d t

\int in (x) d x

n = 1 \sum \infty \frac{1}{( 2 n - 1 ) ^{5}}

\frac{d}{d x} (4 x^{5} - 5 x^{4})

What is the derivative of f(x)=(-3x+1)/(x^2-x) ?
The derivative of f(x)=(-3x+1)/(x^2-x) is (3x^2-2x+1)/((x^2-x)^2)
What is the first derivative of f(x)=(-3x+1)/(x^2-x) ?
The first derivative of f(x)=(-3x+1)/(x^2-x) is (3x^2-2x+1)/((x^2-x)^2)