What is the derivative of f(x)=(-2x)/(x^5-x-1) ?

The derivative of f(x)=(-2x)/(x^5-x-1) is -(2(-4x^5-1))/((x^5-x-1)^2)

What is the first derivative of f(x)=(-2x)/(x^5-x-1) ?

The first derivative of f(x)=(-2x)/(x^5-x-1) is -(2(-4x^5-1))/((x^5-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) = \frac{- 2 x}{x ^{5} - x - 1}

- \frac{2 ( - 4 x ^{5} - 1 )}{( x ^{5} - x - 1 ) ^{2}}

Solution steps

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

Simplify

\frac{- 2 x}{x ^{5} - x - 1} : - \frac{2 x}{x ^{5} - x - 1}

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

\frac{d}{d x} (x^{5} - x - 1) = 5 x^{4} - 1

= - 2 \cdot \frac{1 \cdot ( x ^{5} - x - 1 ) - ( 5 x ^{4} - 1 ) x}{( x ^{5} - x - 1 ) ^{2}}

Simplify

- 2 \cdot \frac{1 \cdot ( x ^{5} - x - 1 ) - ( 5 x ^{4} - 1 ) x}{( x ^{5} - x - 1 ) ^{2}} : - \frac{2 ( - 4 x ^{5} - 1 )}{( x ^{5} - x - 1 ) ^{2}}

= - \frac{2 ( - 4 x ^{5} - 1 )}{( x ^{5} - x - 1 ) ^{2}}

\int yz e^{x y} d x

y = x, (4, 2)

\int \frac{4 x ^{2} + 5 x + 4}{( x ^{2} + 1 ) ^{2}} d x

\int 2 sin (x) + sin (2 x) d x

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

What is the derivative of f(x)=(-2x)/(x^5-x-1) ?
The derivative of f(x)=(-2x)/(x^5-x-1) is -(2(-4x^5-1))/((x^5-x-1)^2)
What is the first derivative of f(x)=(-2x)/(x^5-x-1) ?
The first derivative of f(x)=(-2x)/(x^5-x-1) is -(2(-4x^5-1))/((x^5-x-1)^2)