Calculus and Its Applications, Chapter 1, Review Exercises, Section Review Exercises, Problem 48

Differentiate. For $\displaystyle y = \frac{3}{42}x^7 - 10x^3 + 13x^2 28x - 2$. Find $y''$


$\begin{equation} \begin{aligned} y' &= \frac{3}{42} \cdot \frac{d}{dx} (x^7) - 10 \cdot \frac{d}{dx} (x^3) + 13 \cdot \frac{d}{dx} (x^2) + 28 \cdot \frac{d}{dx} (x) - \frac{d}{dx} (2)\\ \\ y' &= \frac{3}{42} \cdot 7x^{7 - 1} - 10 \cdot 3x^{3 - 1} + 13 \cdot 2x^{2 - 1} + 28(1) - 0 \\ \\ y' &= \frac{1}{2} x^6 - 30x^2 + 26x + 28 \end{aligned} \end{equation}$


Then,

$\begin{equation} \begin{aligned} y'' &= \frac{1}{2} \cdot \frac{d}{dx} (x^6) - 30 \cdot \frac{d}{dx} (x^2) + 26 \cdot \frac{d}{dx} (x) + \frac{d}{dx} (28)\\ \\ y'' &= \frac{1}{2} \cdot 6x^{6 - 1} - 30 \cdot 2 x^{2 - 1} + 26(1) + 0\\ \\ y'' &= 3x^5 - 60x + 26 \end{aligned} \end{equation}$