What is the derivative of y=ln((9+e^x)/(9-e^x)) ?

The derivative of y=ln((9+e^x)/(9-e^x)) is (18e^x)/((e^x+9)(-e^x+9))

What is the first derivative of y=ln((9+e^x)/(9-e^x)) ?

The first derivative of y=ln((9+e^x)/(9-e^x)) is (18e^x)/((e^x+9)(-e^x+9))

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{9 + e ^{x}}{9 - e ^{x}})

\frac{18 e ^{x}}{( e ^{x} + 9 ) ( - e ^{x} + 9 )}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{\frac{9 + e ^{x}}{9 - e ^{x}}} \cdot \frac{18 e ^{x}}{( 9 - e ^{x} ) ^{2}} : \frac{18 e ^{x}}{( e ^{x} + 9 ) ( - e ^{x} + 9 )}

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

x \to 3 lim (\frac{x + 3}{x ^{2} - 9})

\frac{2 s + 2}{s + 10}

x \to 1 - lim (\frac{- 1}{( x - 1 ) ^{\frac{2}{3}}})

\frac{\partial}{\partial y} (3 x^{2} - 2 y^{3} + 6 x y^{2} - 8)

x^{2} 1, 8

What is the derivative of y=ln((9+e^x)/(9-e^x)) ?
The derivative of y=ln((9+e^x)/(9-e^x)) is (18e^x)/((e^x+9)(-e^x+9))
What is the first derivative of y=ln((9+e^x)/(9-e^x)) ?
The first derivative of y=ln((9+e^x)/(9-e^x)) is (18e^x)/((e^x+9)(-e^x+9))