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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d x} (1) = 0

= \frac{1}{2} ln (\frac{1 + x}{1 - x}) + \frac{x}{( x + 1 ) ( - x + 1 )} - 0

Simplify

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

8 sin (2 x)

\frac{d}{d x} (\frac{1 + x}{x ^{4}})

\frac{\partial}{\partial x} (5 y^{4} cos (3 x))

\frac{d}{d x} (\frac{4 x ^{3} + 6}{x})

\int_{25}^{50} 15 d x

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