Question

find the derivative of the following function by first expanding the expression
$g(r)=(3r^{3}+7r + 4)(r^{2}+8)$
$g(r)=square$

Explanation:

Step1: Expand the function

\[

$$\begin{align*} g(r)&=(3r^{3}+7r + 4)(r^{2}+8)\\ &=3r^{3}\times r^{2}+3r^{3}\times8+7r\times r^{2}+7r\times8 + 4\times r^{2}+4\times8\\ &=3r^{5}+24r^{3}+7r^{3}+56r + 4r^{2}+32\\ &=3r^{5}+31r^{3}+4r^{2}+56r + 32 \end{align*}$$

\]

Step2: Differentiate term - by - term

The power rule for differentiation is $\frac{d}{dr}(r^{n})=nr^{n - 1}$.
\[

$$\begin{align*} g^{\prime}(r)&=\frac{d}{dr}(3r^{5})+\frac{d}{dr}(31r^{3})+\frac{d}{dr}(4r^{2})+\frac{d}{dr}(56r)+\frac{d}{dr}(32)\\ &=3\times5r^{4}+31\times3r^{2}+4\times2r+56\times1+0\\ &=15r^{4}+93r^{2}+8r + 56 \end{align*}$$

\]

Answer:

$15r^{4}+93r^{2}+8r + 56$