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

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

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

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

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derivative of

y = 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 )}

f (x) = (ln (x))^{3}, at x = 9

x \to 0 lim (\frac{2 - 2 cos ( x )}{x})

\int_{a}^{b} x d x

y^{2} - 2 x = 9, x - y = 3 [- 4.5, 8]

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