If $$(cosx)y=(siny)x$$, then $$dydx=$$

$$(cosx)y=(siny)xy$$.$$log(cosx)=x$$.$$log(siny)y.1cosx($$−$$sinx)+y$$'.$$log(cosx)=xcosysiny$$.y'$$+log(siny)y$$|[logcosx−x.cosysiny]$$=y$$.$$sinxcosx+log(siny)y$$|[logcosx−xcoty]$$=y$$.$$tanx+log(siny)y$$|$$=logsiny+ytanxlogcosx$$−x.coty

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If ${(\cos x)}^{y} = {(\sin y)}^{x}$ , then $\frac{d y}{d x} =$

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\frac{logsin y + y \tan x}{logcos x - x \cot y}

\frac{logsin y - y \tan x}{logcos x + \cot y}

\frac{logsin y}{logcos x}

\frac{logcos x}{logsin y}

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Correct option is A

${(\cos x)}^{y} = {(\sin y)}^{x}$

$y . \log (\cos x) = x . \log (\sin y)$

$y . \frac{1}{\cos x} (- \sin x) + y' . \log (\cos x) = x \frac{\cos y}{\sin y} . y' + \log (\sin y)$

$y^{|} [logcos x - x . \frac{\cos y}{\sin y}] = y . \frac{\sin x}{\cos x} + \log (\sin y)$

$y^{|} [logcos x - x \cot y] = y . \tan x + \log (\sin y)$

$y^{|} = \frac{logsin y + y \tan x}{logcos x - x . \cot y}$

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If (cosx)y=(siny)x, then dydx=

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If ${(\cos x)}^{y} = {(\sin y)}^{x}$ , then $\frac{d y}{d x} =$