What is the derivative of y=ln(e^{-x}+xe^{-x}) ?

The derivative of y=ln(e^{-x}+xe^{-x}) is -x/(1+x)

What is the first derivative of y=ln(e^{-x}+xe^{-x}) ?

The first derivative of y=ln(e^{-x}+xe^{-x}) is -x/(1+x)

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derivative of

y = ln (e^{- x} + x e^{- x})

- \frac{x}{1 + x}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= - \frac{x}{1 + x}

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What is the derivative of y=ln(e^{-x}+xe^{-x}) ?
The derivative of y=ln(e^{-x}+xe^{-x}) is -x/(1+x)
What is the first derivative of y=ln(e^{-x}+xe^{-x}) ?
The first derivative of y=ln(e^{-x}+xe^{-x}) is -x/(1+x)