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`(sinx)^(logx)` |
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Answer» Let `y = (sinx)^logx` Taking logs both sides, `logy = logx logsin x ` Differentiating both sides w.r.t. x `1/y dy/dx = logx(1/sinx)(cosx)+ logsinx (1/x)` `=>1/ydy/dx = logxcotx+logsinx/x` `=> dy/dx = y( logxcotx+logsinx/x)` `=> dy/dx = (sinx)^logx( logxcotx+logsinx/x)`, which is the required solution. |
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