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

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

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

The first derivative of 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 >

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

\frac{d}{d x} (e^{x^{2}} 2 x)

\int \frac{sin ( 2 x )}{2 cos ( x )} d x

\int x sin (x) d x

\frac{\partial}{\partial x} (cos (x y) x)

(3 x^{2} - 12)^{^{'}}

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