What is the derivative of y=ln(sqrt((x+1)/(x-1))) ?

The derivative of y=ln(sqrt((x+1)/(x-1))) is -1/((x-1)(x+1))

What is the first derivative of y=ln(sqrt((x+1)/(x-1))) ?

The first derivative of y=ln(sqrt((x+1)/(x-1))) is -1/((x-1)(x+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{\frac{x + 1}{x - 1}} (- \frac{1}{x + 1 ( x - 1 ) ^{\frac{3}{2}}})

Simplify

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

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

\frac{d y}{d t} = 2 t y^{2} + 3 y^{2}

x \to + 0 + lim ((7 x)^{x})

\int_{- \infty}^{\infty} x e^{x^{2}} d x

\frac{d y}{d x} = - k (y - a)

3 y ln (x) - x y^{^{'}} = 0

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