Answer
a) $f'(x) = \frac{(x-1)(15x+1)}{2\sqrt{3x+1}}$
b) $f'(x) = \frac{-3 * (3x+1)^2 (3x+2)}{x^7}$
Work Step by Step
Using the rules of derivatives, we find:
a) $f'(x) = [(3x+1)^{\frac{1}{2}}]'[(x-1)^2] +[(3x+1)^{\frac{1}{2}}][(x-1)^2]' = \frac{1}{2} * (3x+1)^{-\frac{1}{2}} * (x-1)^2 + (3x+1)^{\frac{1}{2}} * 2(x-1) = \frac{(x-1)(15x+1)}{2\sqrt{3x+1}}$
b) $f'(x) = 3 * (\frac{3x+1}{x^2})^2 * [\frac{3x+1}{x^2}]' = 3 * (\frac{3x+1}{x^2})^2 * \frac{3*x^2 - 2x * (3x+1)}{x^4} = \frac{-3 * (3x+1)^2 (3x+2)}{x^7}$
