What is the derivative of ln(xe^{-2x}) ?

The derivative of ln(xe^{-2x}) is 1/x-2

What is the first derivative of ln(xe^{-2x}) ?

The first derivative of ln(xe^{-2x}) is 1/x-2

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derivative of

ln (x e^{- 2 x})

\frac{1}{x} - 2

Solution steps

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

Simplify

ln (x e^{- 2 x}) : ln (x) - 2 x

= \frac{d}{d x} (ln (x) - 2 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (ln (x)) - \frac{d}{d x} (2 x)

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

\frac{d}{d x} (2 x) = 2

= \frac{1}{x} - 2

\int \frac{x - 2}{x + 2} d x

y = x^{3} - 3 x + 5, (3, 23)

\frac{d}{d t} (5 sin (t) cos (t))

\int 8 ln (3 x) d x

\int_{1}^{2} ln (3 x) d x

What is the derivative of ln(xe^{-2x}) ?
The derivative of ln(xe^{-2x}) is 1/x-2
What is the first derivative of ln(xe^{-2x}) ?
The first derivative of ln(xe^{-2x}) is 1/x-2