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}
$