What is (\partial)/(\partial x)(1/(1-e^{-x)}) ?

The answer to (\partial)/(\partial x)(1/(1-e^{-x)}) is -(e^{-x})/((1-e^{-x))^2}

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(\partial)/(\partial x)(1/(1-e^{-x)})

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 0 lim (x cot (\frac{10}{3} x))

y = x^{2} 8 x - 9

y^{^{''}} + 4 y = sec^{3} (2 x)

y = arctan (- x), at x = 3

\int 2 - x^{2} d x

What is (\partial)/(\partial x)(1/(1-e^{-x)}) ?
The answer to (\partial)/(\partial x)(1/(1-e^{-x)}) is -(e^{-x})/((1-e^{-x))^2}