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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

x = 0 \sum \infty sin (x)

\int x x - 2 d x

\frac{e ^{x}}{cos ( x )}

\frac{d}{d x} (4 sin (x) cos (x))

\int \frac{2}{5} x d x

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