Answer
$\frac{dy}{dx}=2x^3-3x-1$
$\frac{d^2y}{dx^2}=6x^2-3$
$\frac{d^3y}{dx^3}=12x$
$\frac{d^4y}{dx^4}=12$
$\frac{d^n y}{dx^n}=0$ for $n\geq5$
Work Step by Step
$y=\frac{x^4}{2}-\frac{3}{2}x^2-x$
Find the first derivative
$\frac{dy}{dx}=\frac{1}{2}\frac{d}{dy}(x^4)-\frac{3}{2}\frac{d}{dy}(x^2)-\frac{d}{dy}(x)$
Using the Power Rule:
$\frac{dy}{dx}=\frac{1}{2}(4)x^{4-1}-\frac{3}{2}(2)x^{2-1}-(1)x^{1-1}$
$\frac{dy}{dx}=2x^3-3x-1$
Find the second derivative:
$\frac{d^2y}{dx^2}=2\frac{d}{dx}(x^3)-3\frac{d}{dx}(x)-\frac{d}{dx}(1)$
Using the Power Rule:
$\frac{d^2y}{dx^2}=2(3)x^{3-1}-3(1)x^{1-1}-(0)$
$\frac{d^2y}{dx^2}=6x^2-3$
Find the third derivative:
$\frac{d^3y}{dx^3}=6\frac{d}{dx}(x^2)-\frac{d}{dx}(3)$
Using the Power Rule:
$\frac{d^3y}{dx^3}=6(2)x^{2-1}-(0)$
$\frac{d^3y}{dx^3}=12x$
Find the fourth derivative:
$\frac{d^4y}{dx^4}=12\frac{d}{dx}(x)$
Using the Power Rule:
$\frac{d^4y}{dx^4}=12(1)x^{1-1}$
$\frac{d^4y}{dx^4}=12$
Find the fifth derivative
$\frac{d^5 y}{dx^5}=\frac{d}{dx}(12)$
$\frac{d^5y}{dx^5}=0$
and the $n$ derivative for $n\geq$5 will be always 0
$\frac{d^n y}{dx^n}=0$
