If x^y=y^{sinx}(tanx)^{cosx}, then (logx-{sinx}/y)\frac{dy}{dx}=

Correct Answer is : cosxlogy−sinxlog(tanx)+cosecx−yx

\cos\,x\,\log\,y-\sin\,x\,\log\,(tanx)+cosec\,x-y/x

\cos\,x\,\log\,y-\sin\,x\,\log\,(tanx)+\cos^2x\,cosec\,x-y/x

TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 65 of 160

Marks: +1, -0

If xy=ysinx(tanx)cosx, then (logx−

sinx

)

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