Science:Math Exam Resources/Courses/MATH100/December 2011/Question 05 (c)/Solution 1

From the initial form of the derivative, as computed in part (a),

$f'(x)={\frac {6}{5}}x^{-4/5}+{\frac {6}{5}}x^{1/5}$

we find that the second derivative is

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