[tex]f(x) = \sqrt[3]{6x+1} = ( 6x +1 )^{\tfrac 13}\\\\\\f'(x) = \dfrac 13 (6x +1)^{\tfrac 13 - 1} \cdot (6+0) = 2 (6x+1)^{-\tfrac{2}3}\\\\\\f''(x) = 2 \left( -\dfrac 23 \right) (6x+1)^{-\tfrac 23 - 1} \cdot(6+0)= -8(6x+1)^{-\tfrac 53}\\\\\\f'''(x) = -8 \left( - \dfrac 53\right) (6x+1)^{-\tfrac 53 - 1} \cdot (6+0) = 80(6x+1)^{-\tfrac 83}\\\\\text{Hence the third derivative of f(x) is}~ 80(6x+1)^{-\tfrac 83}\\[/tex]
