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

The derivative of f(x)=(x^2+2x+1)/(x^2-2x+1) is -(4(x+1))/((x-1)^3)

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

The first derivative of f(x)=(x^2+2x+1)/(x^2-2x+1) is -(4(x+1))/((x-1)^3)

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{x ^{2} + 2 x + 1}{x ^{2} - 2 x + 1}

- \frac{4 ( x + 1 )}{( x - 1 ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= - \frac{4 ( x + 1 )}{( x - 1 ) ^{3}}

\int 7 sin (5 x) sin (x) d x

x \to 0 lim (4 x - ∣ x ∣)

\int - x e^{\frac{- x ^{2}}{2}} d x

x^{^{''}} - 5 x^{^{'}} + 6 x = cos (t) \cdot e^{- t}

\frac{\partial}{\partial x} (\frac{u ^{2}}{x ^{2}})

What is the derivative of f(x)=(x^2+2x+1)/(x^2-2x+1) ?
The derivative of f(x)=(x^2+2x+1)/(x^2-2x+1) is -(4(x+1))/((x-1)^3)
What is the first derivative of f(x)=(x^2+2x+1)/(x^2-2x+1) ?
The first derivative of f(x)=(x^2+2x+1)/(x^2-2x+1) is -(4(x+1))/((x-1)^3)