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differentiate |
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Answer» y= (√x)^y take log log y = y log√x differentiate both side wrtx 1/y. y' = y'log√x + 1/2√x. 1/√x .y 1/y. y'= y' log√x + 1/2x . y y'( 1/y - log √x) = y/2x y'= y/2x( 1/y - log √x) |
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