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

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

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

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

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derivative of

f (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}}

\frac{d y}{d x} = e^{(3 x + 2 y)}

f (x) = x^{2} + 3 x, at x = 1

\frac{d}{d x} (4 - \frac{15}{4 x ^{\frac{1}{4}}})

\frac{d}{d x} (x^{3} + 3 x^{2} - 8)

\int 3 cos^{3} (2 x) d x

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