What is the derivative of (e^{2x}/(1+e^{2x)}) ?

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (e^{2 x}) = e^{2 x} \cdot 2

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

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

Simplify

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

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

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