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JEE MainMathematicsEasyDifferentiation

If y=log⁡e(sinx),y = \log_{\sqrt{e}}(sinx), then dydx\frac{dy}{dx} is

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y=11/2log⁡e(sinx)=2ln(sinx)y = \frac{1}{1/2}\log_{e}(sinx) = 2ln(sinx)
dydx=2cosxsinx=2cotx\frac{dy}{dx} = 2\frac{cosx}{sinx} = 2cotx

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx

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