What is the derivative of ${log}_{e} (x)$ $log_e(x)$ ?

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Answer 1 · 2018-04-04T04:36:05

$\frac{1}{x}$ $1/x$

${log}_{e} (x)$ $log_e(x)$ is commonly denoted as $ln (x)$ $ln(x)$ , the natural log.

$\Rightarrow \frac{d}{d x} ln (x) = \frac{1}{x}$ $=>d/(dx) ln(x) = 1/x$

If you would like a proof, we can derive it from the limit definition:

$lim_(delta x->0)(f(x+delta x)-f(x))/(delta x)$

$= lim_(delta x->0)(ln(x+delta x)-ln(x))/(delta x)$

$= lim_(delta x->0)(ln((x+delta x)/(x)))/(delta x)$

$= lim_(delta x->0)1/(delta x)ln(1+(delta x)/x)$

$= lim_(delta x->0)ln((1+(delta x)/x)^(1/(delta x)))$

$"Let " tau equiv (delta x)/x$ :

$= lim_(delta tau->0)ln((1+tau)^(1/(xtau)))$

$= lim_(delta tau->0)ln[((1+tau)^(1/(tau)))^(1/x)]$

$= ln[(e)^(1/x)]$

$= 1/x ln(e)$

$= 1/x (1)$

$= 1/x$

What is the derivative of log_e(x)loge(x)log_e(x)?