What is the derivative of ln((8-x)/(8+x)) ?

The derivative of ln((8-x)/(8+x)) is -(16)/((-x+8)(x+8))

What is the first derivative of ln((8-x)/(8+x)) ?

The first derivative of ln((8-x)/(8+x)) is -(16)/((-x+8)(x+8))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

ln (\frac{8 - x}{8 + x})

- \frac{16}{( - x + 8 ) ( x + 8 )}

Solution steps

\frac{d}{d x} (ln (\frac{8 - x}{8 + x}))

Apply the chain rule:

\frac{1}{\frac{8 - x}{8 + x}} \frac{d}{d x} (\frac{8 - x}{8 + x})

= \frac{1}{\frac{8 - x}{8 + x}} \frac{d}{d x} (\frac{8 - x}{8 + x})

\frac{d}{d x} (\frac{8 - x}{8 + x}) = - \frac{16}{( 8 + x ) ^{2}}

= \frac{1}{\frac{8 - x}{8 + x}} (- \frac{16}{( 8 + x ) ^{2}})

Simplify

\frac{1}{\frac{8 - x}{8 + x}} (- \frac{16}{( 8 + x ) ^{2}}) : - \frac{16}{( - x + 8 ) ( x + 8 )}

= - \frac{16}{( - x + 8 ) ( x + 8 )}

72 x - \frac{4 x ^{3}}{3}

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

\int \frac{1}{x + 2} d x

\frac{d}{d θ} (2 cos (θ))

\frac{d y}{d x} = (4 x^{2} + 3 x + 2) (y + 5)

What is the derivative of ln((8-x)/(8+x)) ?
The derivative of ln((8-x)/(8+x)) is -(16)/((-x+8)(x+8))
What is the first derivative of ln((8-x)/(8+x)) ?
The first derivative of ln((8-x)/(8+x)) is -(16)/((-x+8)(x+8))