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

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

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

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

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derivative of

\frac{e ^{x}}{e ^{x} + 1}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

y^{^{''}} + 5 y = - 180 x^{2} e^{5 x}

y^{^{'}} = \frac{( 4 t + y )}{( 4 y - t )}

x^{5} + 2, x + 2

\int sin (m x) d x

\frac{d y}{d x} = (\frac{y - x + 1}{y - x - 6})

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