What is the derivative of ln((4+e^x/(4-e^x))) ?

The derivative of ln((4+e^x/(4-e^x))) is (8e^x)/((e^x+4)(-e^x+4))

What is the first derivative of ln((4+e^x/(4-e^x))) ?

The first derivative of ln((4+e^x/(4-e^x))) is (8e^x)/((e^x+4)(-e^x+4))

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Popular >

\frac{d}{d x} (ln (\frac{4 + e ^{x}}{4 - e ^{x}}))

\frac{8 e ^{x}}{( e ^{x} + 4 ) ( - e ^{x} + 4 )}

Solution steps

\frac{d}{d x} (ln (\frac{4 + e ^{x}}{4 - e ^{x}}))

Apply the chain rule:

\frac{1}{\frac{4 + e ^{x}}{4 - e ^{x}}} \frac{d}{d x} (\frac{4 + e ^{x}}{4 - e ^{x}})

= \frac{1}{\frac{4 + e ^{x}}{4 - e ^{x}}} \frac{d}{d x} (\frac{4 + e ^{x}}{4 - e ^{x}})

\frac{d}{d x} (\frac{4 + e ^{x}}{4 - e ^{x}}) = \frac{8 e ^{x}}{( 4 - e ^{x} ) ^{2}}

= \frac{1}{\frac{4 + e ^{x}}{4 - e ^{x}}} \cdot \frac{8 e ^{x}}{( 4 - e ^{x} ) ^{2}}

Simplify

\frac{1}{\frac{4 + e ^{x}}{4 - e ^{x}}} \cdot \frac{8 e ^{x}}{( 4 - e ^{x} ) ^{2}} : \frac{8 e ^{x}}{( e ^{x} + 4 ) ( - e ^{x} + 4 )}

= \frac{8 e ^{x}}{( e ^{x} + 4 ) ( - e ^{x} + 4 )}

\int (\frac{x ^{3}}{3}) (\frac{1}{x}) d x

\frac{d y}{d x} = \frac{y}{x} + 8 x + 1

\int \frac{1}{( x - 43 ) ^{0.98}} d x

\int cos (7 x - 1) d x

\frac{d}{d x} ((2 x + 3) (x^{2} - 5))

What is the derivative of ln((4+e^x/(4-e^x))) ?
The derivative of ln((4+e^x/(4-e^x))) is (8e^x)/((e^x+4)(-e^x+4))
What is the first derivative of ln((4+e^x/(4-e^x))) ?
The first derivative of ln((4+e^x/(4-e^x))) is (8e^x)/((e^x+4)(-e^x+4))