If f(x)=xf(x) = \sqrt{x}f(x)=x, then f′(x)=f'(x) =f′(x)=Ax/2\sqrt x/2x/2B2x2\sqrt x2xC1x\tfrac{1}{\sqrt x}x1D12x\tfrac{1}{2\sqrt x}2x1check_circleExplanationx=x1/2\sqrt x = x^{1/2}x=x1/2. f′(x)=12x−1/2=1/(2x)f'(x) = \tfrac{1}{2} x^{-1/2} = 1/(2\sqrt x)f′(x)=21x−1/2=1/(2x).