Answer
a) $g'(\frac{\pi}{3}) = 40\sqrt{3}$
b) h'(2) = 7500
Work Step by Step
We find:
a) $g'(x) = f'(sec(x)) * sec(x)*tan(x)$
$g'(\frac{\pi}{3}) = f'(sec(\frac{\pi}{3})) * sec(\frac{\pi}{3})*tan(\frac{\pi}{3}) = f'(2) * 2 * \sqrt{3} = (2 * 2 * f(2)) * 2 * \sqrt{3} = (4*5)*2*\sqrt{3} = 40\sqrt{3}$
b) $h'(x) = 4 (\frac{f(x)}{(x-1)} )^3 * \frac{f'(x)*(x-1) - f(x)}{(x-1)^2}$
$h'(2) = 4 (\frac{f(2)}{(2-1)} )^3 * \frac{f'(2)*(2-1) - f(2)}{(2-1)^2} = 4 * (f(2))^3 * (f'(2) - f(2)) = 4 * (5)^3 * (2*2*f(2)-5) = 4 * 125 * (15) = 7500$
