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    If y=log√e(sinx)y = log_{√e}(sin x)y=log√e​(sinx) then dydx\frac{dy}{dx}dxdy​ is :​
    Question

    If y=loge(sinx)y = log_{√e}(sin x) then dydx\frac{dy}{dx} is :​

    A.

    ecotx\sqrt e \cot x \\​​

    B.

    1ecotx\frac{1}{\sqrt e} \cot x \\​​

    C.

    2cotx2 \cot x \\​​

    D.

    12cotx\frac{1}{2} \cot x​​

    Correct option is C

    Given:
    y=loge(sinx)y = \log_{\sqrt{e}}(\sin x)​​
    Concept used:
    logab=lnblna,andddx(ln(sinx))=cotx\log_a b = \frac{\ln b}{\ln a}, and \frac{d}{dx}(\ln(\sin x)) = \cot x​​
    Solution:
    y=ln(sinx)ln(e) ln(e)=ln(e1/2)=12y = \frac{\ln(\sin x)}{\ln(\sqrt{e})} \\ \ \\\ln(\sqrt{e}) = \ln(e^{1/2}) = \frac{1}{2}​​
    So,
    y=ln(sinx)1/2=2ln(sinx)y = \frac{\ln(\sin x)}{1/2} = 2\ln(\sin x)​​
    Differentiate:
    dydx=2×cosxsinx=2cotx\frac{dy}{dx} = 2 \times \frac{\cos x}{\sin x} = 2\cot x​​
    Correct answer is (c) 2cotx2 \cot x \\​​

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