What is the derivative of ln(xe^{-9x}) ?

The derivative of ln(xe^{-9x}) is 1/x-9

What is the first derivative of ln(xe^{-9x}) ?

The first derivative of ln(xe^{-9x}) is 1/x-9

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\frac{d}{d x} (ln (x e^{- 9 x}))

\frac{1}{x} - 9

Solution steps

\frac{d}{d x} (ln (x e^{- 9 x}))

Simplify

ln (x e^{- 9 x}) : ln (x) - 9 x

= \frac{d}{d x} (ln (x) - 9 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (ln (x)) - \frac{d}{d x} (9 x)

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

\frac{d}{d x} (9 x) = 9

= \frac{1}{x} - 9

\frac{\partial}{\partial x} (\frac{c - b}{a - x})

\frac{d}{d x} (m x^{n} - n x^{m})

y^{^{'}} = \frac{( 1 + x )}{x y ^{14}}

y = (x^{2} - 15)^{4}, at x = 4

\frac{d}{d x} (ln (- cos (x^{2}) + 1))

What is the derivative of ln(xe^{-9x}) ?
The derivative of ln(xe^{-9x}) is 1/x-9
What is the first derivative of ln(xe^{-9x}) ?
The first derivative of ln(xe^{-9x}) is 1/x-9