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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x + y^{2} = 20, x + y = 0

f (x) = x^{2} - 3 x + 4

x \to - 8 lim (4 x - 2)

\frac{\partial}{\partial y} (\frac{x}{( y + z )})

\frac{\partial}{\partial x} (x ln (x))

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