What is the derivative of ln((e^x/x)) ?

The derivative of ln((e^x/x)) is 1-1/x

What is the first derivative of ln((e^x/x)) ?

The first derivative of ln((e^x/x)) is 1-1/x

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

1 - \frac{1}{x}

Solution steps

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

Simplify

ln (\frac{e ^{x}}{x}) : x - ln (x)

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

Apply the Sum/Difference Rule:

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

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

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

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

= 1 - \frac{1}{x}

\int \frac{1}{e ^{4 x}} d x

\frac{d}{d x} (5 x^{6})

f (x) = \frac{2}{x - 1}, at x = 4

s \to - 1 lim (\frac{1}{s ^{2} + 2 s + 1})

6 - 5 x

What is the derivative of ln((e^x/x)) ?
The derivative of ln((e^x/x)) is 1-1/x
What is the first derivative of ln((e^x/x)) ?
The first derivative of ln((e^x/x)) is 1-1/x