What is the derivative of e+ln(x) ?

The derivative of e+ln(x) is 1/x

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

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

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

\frac{1}{x}

Solution steps

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

Apply the Sum/Difference Rule:

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

= \frac{d e}{d x} + \frac{d}{d x} (ln (x))

\frac{d e}{d x} = 0

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

= 0 + \frac{1}{x}

Simplify

= \frac{1}{x}

x \to 0 lim (3 - 2^{x})

\frac{( 2 s + 8 )}{s ^{2} + 6 s + 25}

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

\int \frac{18 x}{1 - x ^{4}} d x

\frac{1}{s + 1}

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