Question #cab31

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Answer 1 · 2018-01-26T16:42:34

$-cotx. lnx -1/x$

$y = ln (x^sinx)$
$y= sinx lnx$

Differentiating wrt to $cosx$ :

$dy/(d(cosx)) = (dsinx) /(d (cosx)). lnx + (dlnx)/(d(cosx)). sinx$

Differentiating $sinx$ wrt to $cosx$ is equivalent to differenting both wrt to $x$
i.e $((dsinx)/dx )/((dcosx)/dx)$

$dy/(d(cosx)) = cosx /-sinx. lnx + (1/x)/-sinx. sinx$

$dy/(d(cosx)) = -cotx. lnx - 1/x$