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

The derivative of 1/x+ln(x) is -1/(x^2)+1/x

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

The first derivative of 1/x+ln(x) is -1/(x^2)+1/x

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

- \frac{1}{x ^{2}} + \frac{1}{x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x ^{2}}

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

= - \frac{1}{x ^{2}} + \frac{1}{x}

5 x^{2} - x^{3}

f (x) = \frac{1}{x ^{2} + 4}

f (x) = \frac{5 x}{x ^{2} - 4}, at x = 3

x \to 6 lim (x - 6)

\int - 2 tan (x) d x

What is the derivative of 1/x+ln(x) ?
The derivative of 1/x+ln(x) is -1/(x^2)+1/x
What is the first derivative of 1/x+ln(x) ?
The first derivative of 1/x+ln(x) is -1/(x^2)+1/x