Question

how it is solve......dy/dx of cosx/logx

Answer 1

here,
$y=\frac{cosx}{logx}$
we know, the concept of defferentiation
e.g.,
$if\:\:y= \frac{f(x)}{g(x)} \\ \\ \\ then \: \: \: \: \frac{dy}{dx} = \frac{g(x) \frac{df(x)}{dx} - f(x) \frac{dg(x)}{dx} }{ {g(x)}^{2} }$
use this concept here,
$so,\frac{dy}{dx}=\frac{logx.\frac{dcosx}{dx}-cosx.\frac{dlogx}{dx}}{logx^2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: = \frac{logx. ( - sinx) - cosx. \frac{1}{x} }{ {(logx)}^{2} }$