What is the derivative of (x^2-4/(x^2-2x)) ?

The derivative of (x^2-4/(x^2-2x)) is -2/(x^2)

What is the first derivative of (x^2-4/(x^2-2x)) ?

The first derivative of (x^2-4/(x^2-2x)) is -2/(x^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (\frac{x ^{2} - 4}{x ^{2} - 2 x})

- \frac{2}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 4}{x ^{2} - 2 x})

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= - \frac{2}{x ^{2}}

\int \frac{2}{x ^{\frac{2}{3}}} d x

2 y \cdot \frac{d y}{d x} = x (y^{2} + 1)

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

y^{^{'}} = 1 - e^{x - y + 8}

\frac{5}{8 s}

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