Diffentiate (sinx)^cosx+(cosx)^tanx with respect to x

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Diffentiate (sinx)^cosx+(cosx)^tanx with respect to x

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Jasmine 1 week 2021-09-07T05:40:50+00:00 1 Answer 0 views 0

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    2021-09-07T05:42:15+00:00

    Answer:

    Given y=〖sinx〗^tanx

    ∴ logy=tanx logsinx .

    Differentiating w.r.t. ‘x’

    1/y dy/dx=d/dx (tanx logsinx)

    =tanx.cosx/sinx+logsinx. sec^2x

    =1+sec^2x.logsinx

    dy/dx=y(1+sec^2x.logsinx).

    Or, dy/dx=〖sinx 〗^tanx (1+sec^2x.logsinx).

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