section 3.3 derivatives and quotients\nquestion 4, 3.3.12\nfind f(x).\nf(x) = (3x + 9)(6x - 5)\nf(x) =…
section 3.3 derivatives and quotients\nquestion 4, 3.3.12\nfind f(x).\nf(x) = (3x + 9)(6x - 5)\nf(x) = \n(type an exact answer.)
Answer
Explanation:
Step1: Apply product - rule
The product - rule states that if $y = u(x)v(x)$, then $y^\prime=u^\prime(x)v(x)+u(x)v^\prime(x)$. Let $u(x)=3x + 9$ and $v(x)=6x - 5$.
Step2: Find $u^\prime(x)$ and $v^\prime(x)$
Differentiate $u(x)$: $u^\prime(x)=\frac{d}{dx}(3x + 9)=3$. Differentiate $v(x)$: $v^\prime(x)=\frac{d}{dx}(6x - 5)=6$.
Step3: Substitute into product - rule
$f^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x)=3(6x - 5)+(3x + 9)\times6$.
Step4: Expand and simplify
[ \begin{align*} f^\prime(x)&=18x-15 + 18x+54\ &=(18x + 18x)+(-15 + 54)\ &=36x + 39 \end{align*} ]
Answer:
$36x + 39$