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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{\frac{x + 1}{x}} (- \frac{1}{x ^{2}}) : - \frac{1}{x ( x + 1 )}

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

\int_{0}^{6} 6 x - x^{2} d x

\int y^{2} - 4 d y

\int cos (x) sin^{4} (x) d x

f (x) = \frac{7}{2} x^{2} - 5 x + 1

θ \to 0 lim (θ cot (5 θ))

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