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

The derivative of (2x^2/(x^2-x)) is -2/((x-1)^2)

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x \to 4 - lim (\frac{- 1}{x - 4})

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

x \to - 4 lim (25)

2 y^{^{''}} + 7 y^{^{'}} - 4 y = 0

6 y^{^{'}} - 2 y = x (y^{4}), y (0) = - 2

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