What is (\partial)/(\partial x)(ln(8-xe^{-x})) ?

The answer to (\partial)/(\partial x)(ln(8-xe^{-x})) is (e^{-x}x-e^{-x})/(8-e^{-x)x}

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(\partial)/(\partial x)(ln(8-xe^{-x}))

\frac{\partial}{\partial x} (ln (8 - x e^{- x}))

\frac{e ^{- x} x - e ^{- x}}{8 - e ^{- x} x}

Solution steps

\frac{\partial}{\partial x} (ln (8 - x e^{- x}))

Apply the chain rule:

\frac{1}{8 - x e ^{- x}} \frac{\partial}{\partial x} (8 - x e^{- x})

= \frac{1}{8 - x e ^{- x}} \frac{\partial}{\partial x} (8 - x e^{- x})

\frac{\partial}{\partial x} (8 - x e^{- x}) = e^{- x} x - e^{- x}

= \frac{1}{8 - x e ^{- x}} (e^{- x} x - e^{- x})

Simplify

\frac{1}{8 - x e ^{- x}} (e^{- x} x - e^{- x}) : \frac{e ^{- x} x - e ^{- x}}{8 - e ^{- x} x}

= \frac{e ^{- x} x - e ^{- x}}{8 - e ^{- x} x}

\frac{d}{d x} ((\frac{4 x ^{2}}{8 - x})^{3})

x \to 1 lim (\frac{x ^{3} + 1}{x - 1})

\frac{\partial}{\partial y} (\frac{y ^{2}}{x + 1})

3 sin (θ)

f (s) = \frac{2}{( 4 s ^{2} - 8 s )}

What is (\partial)/(\partial x)(ln(8-xe^{-x})) ?
The answer to (\partial)/(\partial x)(ln(8-xe^{-x})) is (e^{-x}x-e^{-x})/(8-e^{-x)x}