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

The derivative of (1-e^{2x}/(1+e^{2x)}) is -(4e^{2x})/((1+e^{2x))^2}

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

The first derivative of (1-e^{2x}/(1+e^{2x)}) is -(4e^{2x})/((1+e^{2x))^2}

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

- \frac{4 e ^{2 x}}{( 1 + e ^{2 x} ) ^{2}}

Solution steps

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

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

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

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

Simplify

\frac{( - 2 e ^{2 x} ) ( 1 + e ^{2 x} ) - 2 e ^{2 x} ( 1 - e ^{2 x} )}{( 1 + e ^{2 x} ) ^{2}} : - \frac{4 e ^{2 x}}{( 1 + e ^{2 x} ) ^{2}}

= - \frac{4 e ^{2 x}}{( 1 + e ^{2 x} ) ^{2}}

\int \frac{cos ( x )}{1 - sin ^{2} ( x )} d x

\int \frac{( x - 2 )}{1} d x

\frac{d}{d x} (9 arctan (x + 1 + x^{2}))

x \to 4 lim (\frac{4 x}{x - 4})

\int \frac{sin ( x )}{x ^{2}} d x

What is the derivative of (1-e^{2x}/(1+e^{2x)}) ?
The derivative of (1-e^{2x}/(1+e^{2x)}) is -(4e^{2x})/((1+e^{2x))^2}
What is the first derivative of (1-e^{2x}/(1+e^{2x)}) ?
The first derivative of (1-e^{2x}/(1+e^{2x)}) is -(4e^{2x})/((1+e^{2x))^2}