Question: # \frac{d}{dx} x(1-x)^{\frac{1}{2}} #
Use product rule, # \frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + g'(x)f(x) #, power rule #\frac{d}{dx} [x^n] = nx^{n-1} # and chain rule, # \frac{d}{dx} [f'(g(x))] = f'(g(x)) * g'(x) #
# \frac{d}{dx} x(1-x)^{\frac{1}{2}} = 1(1-x)^{\frac{1}{2}} + \frac{d}{dx}[(1-x)^{\frac{1}{2}}] * x #
# = (1-x)^{\frac{1}{2}} + x * [\frac{1}{2}(1-x)^{\frac{-1}{2}} * -1] #
# = \sqrt{1-x} + x*[\frac{-1}{2} * \frac{1}{\sqrt{1-x}}]#
# = \sqrt{1-x} + \frac{-x}{2\sqrt{1-x}} #
# = \frac{2(1-x)}{2sqrt(1-x)} + \frac{-x}{2\sqrt{1-x}} #
# = \frac{2-3x}{2\sqrt{1-x}} #