What is the derivative of $ln (1 + (\frac{1}{x}))$ $ln(1+(1/x))$ ?

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Answer 1 · 2015-08-30T15:00:28

$\frac{d}{d x} (ln (1 + (\frac{1}{x}))) = \frac{- 1}{x (x + 1)}$ $d/dx(ln(1+(1/x))) = (-1)/(x(x+1))$

Although you could use $\frac{d}{d x} (ln (u)) = \frac{1}{u} \frac{d u}{d x}$ $d/dx (ln(u)) = 1/u (du)/dx$ , the algebra will get messy that way.

Let's rewrite using properties of $ln$ $ln$ .

$y = ln (1 + (\frac{1}{x})) = ln (\frac{x + 1}{x})$ $y = ln(1+(1/x)) = ln((x+1)/x)$

$= ln (x + 1) - ln (x)$ $= ln(x+1) - ln(x)$

So

$\frac{d y}{d x} = \frac{1}{x + 1} - \frac{1}{x} = \frac{- 1}{x (x + 1)}$ $dy/dx = 1/(x+1) - 1/x = (-1)/(x(x+1))$

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