What is the derivative of y=ln((7+e^x)/(7-e^x)) ?

The derivative of y=ln((7+e^x)/(7-e^x)) is (14e^x)/((e^x+7)(-e^x+7))

What is the first derivative of y=ln((7+e^x)/(7-e^x)) ?

The first derivative of y=ln((7+e^x)/(7-e^x)) is (14e^x)/((e^x+7)(-e^x+7))

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derivative of

y = ln (\frac{7 + e ^{x}}{7 - e ^{x}})

\frac{14 e ^{x}}{( e ^{x} + 7 ) ( - e ^{x} + 7 )}

Solution steps

\frac{d}{d x} (ln (\frac{7 + e ^{x}}{7 - e ^{x}}))

Apply the chain rule:

\frac{1}{\frac{7 + e ^{x}}{7 - e ^{x}}} \frac{d}{d x} (\frac{7 + e ^{x}}{7 - e ^{x}})

= \frac{1}{\frac{7 + e ^{x}}{7 - e ^{x}}} \frac{d}{d x} (\frac{7 + e ^{x}}{7 - e ^{x}})

\frac{d}{d x} (\frac{7 + e ^{x}}{7 - e ^{x}}) = \frac{14 e ^{x}}{( 7 - e ^{x} ) ^{2}}

= \frac{1}{\frac{7 + e ^{x}}{7 - e ^{x}}} \cdot \frac{14 e ^{x}}{( 7 - e ^{x} ) ^{2}}

Simplify

\frac{1}{\frac{7 + e ^{x}}{7 - e ^{x}}} \cdot \frac{14 e ^{x}}{( 7 - e ^{x} ) ^{2}} : \frac{14 e ^{x}}{( e ^{x} + 7 ) ( - e ^{x} + 7 )}

= \frac{14 e ^{x}}{( e ^{x} + 7 ) ( - e ^{x} + 7 )}

\int cos^{2} (d) x d x

y^{^{'}} + 0.121 y = - 3

y^{^{'}} = \frac{( 3 x ^{7} e ^{\frac{y}{x}} + x ^{5} y ^{2} )}{x ^{6} y}

\frac{\partial}{\partial x} (\frac{1}{x + c t})

\int_{l n (2)}^{3} (5 e^{x}) d x

What is the derivative of y=ln((7+e^x)/(7-e^x)) ?
The derivative of y=ln((7+e^x)/(7-e^x)) is (14e^x)/((e^x+7)(-e^x+7))
What is the first derivative of y=ln((7+e^x)/(7-e^x)) ?
The first derivative of y=ln((7+e^x)/(7-e^x)) is (14e^x)/((e^x+7)(-e^x+7))