## Thomas' Calculus 13th Edition

$\frac{dy}{dx}=2x^3-3x-1$ $\frac{d^2y}{dx^2}=6x^2-3$ $\frac{d^3y}{dx^3}=12x$ $\frac{d^4y}{dx^4}=12$ $\frac{d^n y}{dx^n}=0$ for $n\geq5$
$y=\frac{x^4}{2}-\frac{3}{2}x^2-x$ 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 $\frac{d^n y}{dx^n}=0$