## Derivatives

Calculate the second derivative of (x^8)-(x^2)+2

Hint
The power rule for the first derivative:
$$\frac{d}{dx}[x^n]=n\cdot x^{n-1}$$$Hint 2 Using the power rule for the first derivative, and applying it twice, we’ll get the second derivative power rule: $$\frac{d^2}{dx^2}[x^n]=\frac{d}{dx}\frac{d}{dx}[x^n]=\frac{d}{dx}[nx^{n-1}]=n\frac{d}{dx}[x^{n-1}]=n(n-1)(x^{n-2})$$$
Using the power rule for the first derivative, and applying it twice, we'll get the second derivative power rule:
$$\frac{d^2}{dx^2}[x^n]=\frac{d}{dx}\frac{d}{dx}[x^n]=\frac{d}{dx}[nx^{n-1}]=n\frac{d}{dx}[x^{n-1}]=n(n-1)(x^{n-2})$$$This calculus problem can be written as: $$\frac{d^2}{dx^2}(x^8-x^2+2)=\frac{d^2}{dx^2}(x^8)-\frac{d^2}{dx^2}(x^2)+\frac{d^2}{dx^2}(2)$$$
Applying the 2nd power rule to the first section of the derivative with $$n=8$$ :
$$\frac{d^2}{dx^2}[x^n]=n(n-1)(x^{n-2})\rightarrow\frac{d^2}{dx^2}(x^8)=56x^6$$$Applying the 2nd power rule to the second section of the derivative with $$n=2$$ : $$\frac{d^2}{dx^2}[x^n]=n(n-1)(x^{n-2})\rightarrow\frac{d^2}{dx^2}(x^2)=2$$$
Applying the constant rule to the third section of the derivative:
$$\frac{d^2}{dx^2}[2x^0]=0$$$Finally, $$\frac{d^{2}}{dx^{2}}(x^{8})-\frac{d^{2}}{dx^{2}}(x^{2})+\frac{d^{2}}{dx^{2}}(2)=56x^6-2$$$
$$56x^6-2$$\$