Answer
$f'(x)=0.005x^4-0.06x^2$
$f''(x)=0.02x^3-0.12x$
Work Step by Step
Apply power rule to the function to find the first derivate
$f'(x)=0.001(5)x^{5-1}-0.02(3)x^{3-1}$
$f'(x)=0.005x^4-0.06x^2$
Apply power rule the first derivate to find the second derivate
$f''(x)=0.005(4)x^{4-1}-0.06(2)x^{2-1}$
$f''(x)=0.02x^3-0.12x$
