Science:Math Exam Resources/Courses/MATH100/December 2011/Question 05 (c)/Solution 2

Using the compact form of the derivative, as computed in part (b),

${\begin{aligned}f'(x)&={\frac {6(x+1)}{5x^{4/5}}}\end{aligned}}$

we find the second derivative by taking the quotient rule:

${\begin{aligned}f''(x)&={\frac {6}{5}}\left({\frac {x^{4/5}-{\frac {4}{5}}x^{-1/5}(x+1)}{x^{8/5}}}\right)\\&={\frac {6x^{-1/5}(x-{\frac {4}{5}}x-{\frac {4}{5}})}{5x^{8/5}}}\\&={\frac {6({\frac {1}{5}}x-{\frac {4}{5}})}{5x^{9/5}}}\\&={\frac {6(x-4)}{25x^{9/5}}}\end{aligned}}$