If  (cosx)y=(siny)x, then  dydx=

# If  ${\left(\mathrm{cos}x\right)}^{y}={\left(\mathrm{sin}y\right)}^{x}$, then  $\frac{dy}{dx}=$

1. A

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

2. B

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

3. C

$\frac{\mathrm{logsin}y}{\mathrm{logcos}x}$

4. D

$\frac{\mathrm{logcos}x}{\mathrm{logsin}y}$

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### Solution:

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

$y.\mathrm{log}\left(\mathrm{cos}x\right)=x.\mathrm{log}\left(\mathrm{sin}y\right)$

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

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

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

${y}^{|}=\frac{\mathrm{logsin}y+y\mathrm{tan}x}{\mathrm{logcos}x-x.\mathrm{cot}y}$  +91

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