II y^{cosx} = x^{siny} , then \frac{dy}{dx} =

Correct Answer is : y(xsinxlogy+siny)x(cosx−ylogxcosy)

\frac{y(xsinxlogy+siny)}{x(cosx-ylogxcosy)}

\frac{y(xsinxlogy-siny)}{x(cosx+ylogxcosy)}

\frac{y(siny-x \log y)}{x(x-ycosy(\log x))}

\frac{y(siny+x \log y)}{x(x+ycosy(\log x))}

TS EAMCET 06 May 2019 Shift 1 Solved Paper

Section: Mathematics

Question : 65 of 160

Marks: +1, -0

II ycosx = xsiny , then

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