What is the derivative of 9k*ln((x+9/(x-9))) ?

The derivative of 9k*ln((x+9/(x-9))) is -(162k)/((x+9)(x-9))

What is the first derivative of 9k*ln((x+9/(x-9))) ?

The first derivative of 9k*ln((x+9/(x-9))) is -(162k)/((x+9)(x-9))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (9 k \cdot ln (\frac{x + 9}{x - 9}))

- \frac{162 k}{( x + 9 ) ( x - 9 )}

Solution steps

\frac{d}{d x} (9 k ln (\frac{x + 9}{x - 9}))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 9 k \frac{d}{d x} (ln (\frac{x + 9}{x - 9}))

Apply the chain rule:

\frac{1}{\frac{x + 9}{x - 9}} \frac{d}{d x} (\frac{x + 9}{x - 9})

= \frac{1}{\frac{x + 9}{x - 9}} \frac{d}{d x} (\frac{x + 9}{x - 9})

\frac{d}{d x} (\frac{x + 9}{x - 9}) = - \frac{18}{( x - 9 ) ^{2}}

= 9 k \frac{1}{\frac{x + 9}{x - 9}} (- \frac{18}{( x - 9 ) ^{2}})

Simplify

9 k \frac{1}{\frac{x + 9}{x - 9}} (- \frac{18}{( x - 9 ) ^{2}}) : - \frac{162 k}{( x + 9 ) ( x - 9 )}

= - \frac{162 k}{( x + 9 ) ( x - 9 )}

y^{^{''}} + b y = 0

- 3 e^{2 t}

x \to - \infty lim (ln (- x))

x \to 3 + lim (\frac{x}{( 3 - x ) ^{2}})

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

What is the derivative of 9k*ln((x+9/(x-9))) ?
The derivative of 9k*ln((x+9/(x-9))) is -(162k)/((x+9)(x-9))
What is the first derivative of 9k*ln((x+9/(x-9))) ?
The first derivative of 9k*ln((x+9/(x-9))) is -(162k)/((x+9)(x-9))