Answer
$f'(x)=2-\frac{15}{4}x^{-1/4}$
$f''(x)=\frac{15}{16}x^{-5/4}$
These functions are reasonable since f' is positive when f is increasing and negative when f is decreasing. The same goes for f'' and f' - since f' is always increasing, f'' is always positive. The slopes of the derivatives are closer to zero wherever the graphs of the original functions appear almost flat.
Work Step by Step
$f(x)=2x-5x^{3/4}$
Use the Power Rule: $f(x) = x^{n},$
then $f'(x) = nx^{n-1}$
$f'(x)=2(1)-5(3/4)x^{-1/4}$
$f'(x)=2-\frac{15}{4}x^{-1/4}$
$f''(x)=0-(\frac{15}{4})(-\frac{1}{4})x^{-5/4}$
$f''(x)=\frac{15}{16}x^{-5/4}$
