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Question #1b978

1 Answer

Dec 4, 2017

$1 - tan x$ $1-tanx$

Explanation:

Assumung $log$ $log$ is in base $e$ $e$ , $log β \equiv ln β$ $logbeta equiv lnbeta$ for the basis of this quesion:

$log (e^{x} cos x)$ $log(e^x cosx )$ = ${log e}^{x} + log (cos x)$ $log e^x + log(cosx)$

$= x + log (cos x)$ $=x+log(cosx)$
$\frac{d}{d x} (x) = 1$ $d/dx ( x) = 1$
$\frac{d}{d x} (log (cos x))$ $d/dx ( log(cosx) )$ ;

We know $d/dx ( log(gamma(x))) = (gamma'(x))/(gamma(x)$

So $d/dx ( log(cosx)) = 1-sinx / cosx = 1-tanx$

This answer differs for $log_10$ but is simple for base $e$

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