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JEE MainMathematicsModerateDifferentiation

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

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y=112log⁡e(sin⁡x)=2ln⁡(sin⁡x)y = \frac{1}{\frac{1}{2}}\log_{e}\left( \sin x \right) = 2\ln\left( \sin x \right)
dydx=2cosxsinx=2cotx\frac{dy}{dx} = 2\frac{cosx}{sinx} = 2cotx

Source: Grass-G / M-19 Mathematics Paper 28 Nov Fourth.docx

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