What is the derivative of ln((3+e^x)/(3-e^x)) ?

The derivative of ln((3+e^x)/(3-e^x)) is (6e^x)/((e^x+3)(-e^x+3))

What is the first derivative of ln((3+e^x)/(3-e^x)) ?

The first derivative of ln((3+e^x)/(3-e^x)) is (6e^x)/((e^x+3)(-e^x+3))

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derivative of

ln (\frac{3 + e ^{x}}{3 - e ^{x}})

\frac{6 e ^{x}}{( e ^{x} + 3 ) ( - e ^{x} + 3 )}

Solution steps

\frac{d}{d x} (ln (\frac{3 + e ^{x}}{3 - e ^{x}}))

Apply the chain rule:

\frac{1}{\frac{3 + e ^{x}}{3 - e ^{x}}} \frac{d}{d x} (\frac{3 + e ^{x}}{3 - e ^{x}})

= \frac{1}{\frac{3 + e ^{x}}{3 - e ^{x}}} \frac{d}{d x} (\frac{3 + e ^{x}}{3 - e ^{x}})

\frac{d}{d x} (\frac{3 + e ^{x}}{3 - e ^{x}}) = \frac{6 e ^{x}}{( 3 - e ^{x} ) ^{2}}

= \frac{1}{\frac{3 + e ^{x}}{3 - e ^{x}}} \cdot \frac{6 e ^{x}}{( 3 - e ^{x} ) ^{2}}

Simplify

\frac{1}{\frac{3 + e ^{x}}{3 - e ^{x}}} \cdot \frac{6 e ^{x}}{( 3 - e ^{x} ) ^{2}} : \frac{6 e ^{x}}{( e ^{x} + 3 ) ( - e ^{x} + 3 )}

= \frac{6 e ^{x}}{( e ^{x} + 3 ) ( - e ^{x} + 3 )}

\frac{d y}{d x} - \frac{1}{x} y = 1 - \frac{1}{x}

d er i v a t i v e f (x) = cos (6 x)

d er i v a t i v e \frac{tan ( 3 x )}{x ^{2}}

\int \frac{24 x ^{6} - 24 x ^{4} + x}{x} d x

x d y = d x

What is the derivative of ln((3+e^x)/(3-e^x)) ?
The derivative of ln((3+e^x)/(3-e^x)) is (6e^x)/((e^x+3)(-e^x+3))
What is the first derivative of ln((3+e^x)/(3-e^x)) ?
The first derivative of ln((3+e^x)/(3-e^x)) is (6e^x)/((e^x+3)(-e^x+3))