What is the derivative of y=ln((5+e^x)/(5-e^x)) ?

The derivative of y=ln((5+e^x)/(5-e^x)) is (10e^x)/((e^x+5)(-e^x+5))

What is the first derivative of y=ln((5+e^x)/(5-e^x)) ?

The first derivative of y=ln((5+e^x)/(5-e^x)) is (10e^x)/((e^x+5)(-e^x+5))

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derivative of

y = ln (\frac{5 + e ^{x}}{5 - e ^{x}})

\frac{10 e ^{x}}{( e ^{x} + 5 ) ( - e ^{x} + 5 )}

Solution steps

\frac{d}{d x} (ln (\frac{5 + e ^{x}}{5 - e ^{x}}))

Apply the chain rule:

\frac{1}{\frac{5 + e ^{x}}{5 - e ^{x}}} \frac{d}{d x} (\frac{5 + e ^{x}}{5 - e ^{x}})

= \frac{1}{\frac{5 + e ^{x}}{5 - e ^{x}}} \frac{d}{d x} (\frac{5 + e ^{x}}{5 - e ^{x}})

\frac{d}{d x} (\frac{5 + e ^{x}}{5 - e ^{x}}) = \frac{10 e ^{x}}{( 5 - e ^{x} ) ^{2}}

= \frac{1}{\frac{5 + e ^{x}}{5 - e ^{x}}} \cdot \frac{10 e ^{x}}{( 5 - e ^{x} ) ^{2}}

Simplify

\frac{1}{\frac{5 + e ^{x}}{5 - e ^{x}}} \cdot \frac{10 e ^{x}}{( 5 - e ^{x} ) ^{2}} : \frac{10 e ^{x}}{( e ^{x} + 5 ) ( - e ^{x} + 5 )}

= \frac{10 e ^{x}}{( e ^{x} + 5 ) ( - e ^{x} + 5 )}

\int x 3 x - 4 d x

\int \frac{x ^{2} + 1}{x ^{3} + 3 x} d x

4 x - 6 y x^{2} + 1 (\frac{d y}{d x}) = 0

n = 1 \sum \infty \frac{1}{n ^{0.08}}

\frac{d}{d x} (a x e^{3 x})

What is the derivative of y=ln((5+e^x)/(5-e^x)) ?
The derivative of y=ln((5+e^x)/(5-e^x)) is (10e^x)/((e^x+5)(-e^x+5))
What is the first derivative of y=ln((5+e^x)/(5-e^x)) ?
The first derivative of y=ln((5+e^x)/(5-e^x)) is (10e^x)/((e^x+5)(-e^x+5))