What is the derivative of ln(xe^{-4x}) ?

The derivative of ln(xe^{-4x}) is 1/x-4

What is the first derivative of ln(xe^{-4x}) ?

The first derivative of ln(xe^{-4x}) is 1/x-4

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derivative of

ln (x e^{- 4 x})

\frac{1}{x} - 4

Solution steps

\frac{d}{d x} (ln (x e^{- 4 x}))

Simplify

ln (x e^{- 4 x}) : ln (x) - 4 x

= \frac{d}{d x} (ln (x) - 4 x)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (ln (x)) - \frac{d}{d x} (4 x)

\frac{d}{d x} (ln (x)) = \frac{1}{x}

\frac{d}{d x} (4 x) = 4

= \frac{1}{x} - 4

\int cos^{10} (x) sin^{3} (x) d x

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\frac{d}{d x} ((2 x - e^{2 x})^{s i n (2 x)})

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What is the derivative of ln(xe^{-4x}) ?
The derivative of ln(xe^{-4x}) is 1/x-4
What is the first derivative of ln(xe^{-4x}) ?
The first derivative of ln(xe^{-4x}) is 1/x-4