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y = x^(1/2) (NB power of '1/2' mean the 'square root'.

Hence

dy/dx = (1/2)x^(-1/2)

or

dy/dx = 1/ [2x^(1/2)]

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lenpollock

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14h ago

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More answers

y = square root of x

y = x(1/2)

y' = (1/2)x(1/2 - 1)

y' = (1/2)x(-1/2)

y' = (1/2)(square root of x)

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Wiki User

14y ago
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