Question

2.4 - 2.6 - using - differentiation - rules: problem (1 point) results for this submission 2 of the answers are not correct. compute the derivative of the given function in two different ways. h(s)=(3s - 10)(2s + 4) a. use the product rule fg=f·g+f·g. (fill in each blank, then simplify.) h(s)=(3s - 10)·(2)+(3)·(2s + 4)=12s - 1: b. use algebra first to simplify h, then differentiate without the product rule. h(s)=6s - 4 note: you can earn partial credit on this problem. preview my answers submit answers your score was recorded. your score was successfully sent to canvas. you have attempted this problem 4 times. you received a score of 40% for this attempt. your overall recorded score is 40%. you have unlimited attempts remaining.

Explanation:

Step1: Apply product - rule

Let \(f(s)=3s - 10\), \(g(s)=2s + 4\). Then \(f'(s)=3\) and \(g'(s)=2\). By the product - rule \((fg)'(s)=f(s)g'(s)+f'(s)g(s)\), so \(h'(s)=(3s - 10)\times2+3\times(2s + 4)\).

Step2: Simplify the product - rule result

\[

$$\begin{align*} h'(s)&=(3s - 10)\times2+3\times(2s + 4)\\ &=6s-20 + 6s+12\\ &=12s - 8 \end{align*}$$

\]

Step3: Simplify \(h(s)\) first

Expand \(h(s)=(3s - 10)(2s + 4)=6s^{2}+12s-20s - 40=6s^{2}-8s - 40\).

Step4: Differentiate the expanded \(h(s)\)

Using the power - rule \((x^n)'=nx^{n - 1}\), \(h'(s)=(6s^{2}-8s - 40)'=12s-8\).

Answer:

a. \(h'(s)=(3s - 10)\cdot(2)+(3)\cdot(2s + 4)=12s - 8\)
b. First expand \(h(s)=6s^{2}-8s - 40\), then \(h'(s)=12s - 8\)