What is the derivative of ln(3xe^{-x}) ?

The derivative of ln(3xe^{-x}) is 1/x-1

What is the first derivative of ln(3xe^{-x}) ?

The first derivative of ln(3xe^{-x}) is 1/x-1

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\frac{d}{d x} (ln (3 x e^{- x}))

\frac{1}{x} - 1

Solution steps

\frac{d}{d x} (ln (3 x e^{- x}))

Simplify

ln (3 x e^{- x}) : ln (3 x) - x

= \frac{d}{d x} (ln (3 x) - x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (ln (3 x)) - \frac{d x}{d x}

\frac{d}{d x} (ln (3 x)) = \frac{1}{x}

\frac{d x}{d x} = 1

= \frac{1}{x} - 1

15 - x^{2}, - 6 x - 1, [- 4, - 1]

\int \frac{9}{( x + 1 ) ^{2}} d x

\int \frac{1 - 2 x}{( x + 1 ) ( x - 5 )} d x

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

y^{^{'}} - e^{y} - 2 x e^{y} = 0

What is the derivative of ln(3xe^{-x}) ?
The derivative of ln(3xe^{-x}) is 1/x-1
What is the first derivative of ln(3xe^{-x}) ?
The first derivative of ln(3xe^{-x}) is 1/x-1