What is the derivative of ln(1+1/(x^2+1)) ?

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

What is the first derivative of ln(1+1/(x^2+1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

ln (1 + \frac{1}{x ^{2} + 1})

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{d}{d x} (1.1 x + 6.3 x^{3})

\frac{d y}{d x} + 3 x y^{2} = 0

\int \frac{3 x ^{4} - 2 x ^{3} - 2 x ^{2} + 2 x - 5}{x ^{3} - x ^{2}} d x

y = \frac{2}{x - 4}, (6, 1)

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