Differentiate (x^0.5)ln(x) with respect to x.

First it's helpful to write f(x) = (x^0.5)ln(x)The product rule is useful here, this may be written in the form (u(x)v(x))' = u(x)v'(x) + u'(x)v(x).Here we will take u(x) = x^0.5 and v(x) = ln(x), meaning f(x) = u(x)v(x). Now, remembering that x^0.5 is simply the square root of x, we find:u'(x) = 0.5x^(-0.5).Differentiating logs can sometimes be tricky, but here we have the simple case of ln(x):v'(x) = (1/x), I would recommend memorising this resultThrough substitution, f'(x) = (x^0.5)(1/x) + (0.5x^(-0.5))(ln(x)) = (x^(-0.5)) + (ln(x))/(2x^0.5) = (2 + ln(x))/(2*x^0.5)