What is the derivative of ln((|x-1|/(|x-3|))) ?

The derivative of ln((|x-1|/(|x-3|))) is -2/((x-1)(x-3))

What is the first derivative of ln((|x-1|/(|x-3|))) ?

The first derivative of ln((|x-1|/(|x-3|))) is -2/((x-1)(x-3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

\frac{d}{d x} (ln (\frac{∣ x - 1 ∣}{∣ x - 3 ∣}))

- \frac{2}{( x - 1 ) ( x - 3 )}

Solution steps

\frac{d}{d x} (ln (\frac{∣ x - 1 ∣}{∣ x - 3 ∣}))

Apply the chain rule:

\frac{1}{\frac{∣ x - 1 ∣}{∣ x - 3 ∣}} \frac{d}{d x} (\frac{∣ x - 1 ∣}{∣ x - 3 ∣})

= \frac{1}{\frac{∣ x - 1 ∣}{∣ x - 3 ∣}} \frac{d}{d x} (\frac{∣ x - 1 ∣}{∣ x - 3 ∣})

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

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

Simplify

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

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

\int 2 x sec^{2} (3 x) x d x

x \to 0 lim (g (x) sin (\frac{1}{x}))

\frac{d}{d x} (sec (x) tan (x))

\frac{\partial}{\partial x} (2 y ln (x))

\int 4 x^{3} e^{- x^{2}} d x

What is the derivative of ln((|x-1|/(|x-3|))) ?
The derivative of ln((|x-1|/(|x-3|))) is -2/((x-1)(x-3))
What is the first derivative of ln((|x-1|/(|x-3|))) ?
The first derivative of ln((|x-1|/(|x-3|))) is -2/((x-1)(x-3))