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}}

Expand

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

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

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

\frac{d ^{2} y}{d x ^{2}} + y = 0

y^{^{'}} - \frac{1}{x} y = x^{2} e^{x}

\frac{d}{d x} (\frac{cos ( 2 x )}{sin ( 3 x )})

\int_{0}^{16} t^{2} + 2 t + 1 d t

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)