What is (\partial)/(\partial x)(-1+ln(x)) ?

The answer to (\partial)/(\partial x)(-1+ln(x)) is 1/x

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(\partial)/(\partial x)(-1+ln(x))

\frac{\partial}{\partial x} (- 1 + ln (x))

\frac{1}{x}

Solution steps

\frac{\partial}{\partial x} (- 1 + ln (x))

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (1) + \frac{\partial}{\partial x} (ln (x))

\frac{\partial}{\partial x} (1) = 0

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

= - 0 + \frac{1}{x}

Simplify

= \frac{1}{x}

\frac{4 x + x ^{2} - 2}{( 2 + x ) ^{2}}

\frac{d}{d x} (\frac{3 x ^{2} - 5}{2 x ^{2} + 1})

x \to - 8 + lim (\frac{( x + 7 )}{x + 8})

\frac{d}{d x} (e^{- a ∣ x ∣})

What is (\partial)/(\partial x)(-1+ln(x)) ?
The answer to (\partial)/(\partial x)(-1+ln(x)) is 1/x