What is the derivative of (1-e^x/(1+e^x)) ?

The derivative of (1-e^x/(1+e^x)) is -(2e^x)/((1+e^x)^2)

What is the first derivative of (1-e^x/(1+e^x)) ?

The first derivative of (1-e^x/(1+e^x)) is -(2e^x)/((1+e^x)^2)

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Popular >

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

- \frac{2 e ^{x}}{( 1 + e ^{x} ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( 1 - e ^{x} ) ( 1 + e ^{x} ) - \frac{d}{d x} ( 1 + e ^{x} ) ( 1 - e ^{x} )}{( 1 + e ^{x} ) ^{2}}

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

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

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

Simplify

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

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

(ln (x^{'}))^{^{'}}

f (x) = 4 sin (2 x) + \frac{2 x ^{5} - x}{x}

\int \frac{5 x ^{2} - 1}{x ^{2}} d x

y = 5 x - x^{2}, (1, 4)

y \to 1 lim (\frac{y ^{2} + 3}{y - 1})

What is the derivative of (1-e^x/(1+e^x)) ?
The derivative of (1-e^x/(1+e^x)) is -(2e^x)/((1+e^x)^2)
What is the first derivative of (1-e^x/(1+e^x)) ?
The first derivative of (1-e^x/(1+e^x)) is -(2e^x)/((1+e^x)^2)