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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

\int x^{2} (ln (x)) d x

\int (\frac{d}{1} (y))^{2} d x

\frac{( 4 s - 5 )}{s ^{2} - 6 s + 13}

n = 1 \sum \infty (\frac{1}{3})^{n}

4 x^{2}, \frac{x}{4}

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