Question

hw 10 - product and quotient rules section 2.5: problem 2 (1 point) compute the derivative of the given function in two different ways. (h(s)=(-7s + 7)(8s + 6)) a) use the product rule, (f(x)g(x)=f(x)cdot g(x)+f(x)cdot g(x)). (fill in each blank, then simplify.) (h(s)=(square)cdot(square)+(square)cdot(square)=square) b) use algebra first to simplify (h), then differentiate without the product rule. (h(s)=square)

Explanation:

Step1: Identify f(s) and g(s) and their derivatives

Let $f(s)=-7s + 7$ and $g(s)=8s + 6$. Then $f'(s)=-7$ and $g'(s)=8$.

Step2: Apply the Product - Rule

By the product - rule $[f(s)g(s)]'=f(s)\cdot g'(s)+f'(s)\cdot g(s)$, we have $h'(s)=(-7s + 7)\cdot8+(-7)\cdot(8s + 6)$.

Step3: Expand and simplify

$h'(s)=-56s+56-56s - 42=-112s + 14$.

Step4: Simplify h(s) first

Expand $h(s)=(-7s + 7)(8s + 6)=-56s^{2}-42s+56s + 42=-56s^{2}+14s + 42$.

Step5: Differentiate the simplified h(s)

Using the power - rule $\frac{d}{ds}(as^{n})=nas^{n - 1}$, we get $h'(s)=-112s+14$.

Answer:

a) $h'(s)=(-7s + 7)\cdot8+(-7)\cdot(8s + 6)=-112s + 14$
b) $h'(s)=-112s + 14$