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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = x^{4} + 6 x^{2} - x

\int (4 x^{3} + x^{2} + 5) d x

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

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

y = cos (x), x = \frac{π}{2}, x = 2 π, y = 0

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