Question

let (f(t)=(t^{2}+5t + 3)(5t^{2}+2)). find (f(t)).
(f(t)=(2t + 5)(5t^{2}+2)+(10t)(t^{2}+5t + 3))
find (f(2)).
(f(2)=)

Explanation:

Step1: Identify the product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Here, let $u=(t^{2}+5t + 3)$ and $v=(5t^{2}+2)$. Then $u' = 2t+5$ and $v'=10t$. So $f'(t)=(2t + 5)(5t^{2}+2)+(10t)(t^{2}+5t + 3)$.

Step2: Substitute $t = 2$

First, calculate $(2t + 5)$ when $t = 2$: $2\times2+5=4 + 5=9$.
Second, calculate $(5t^{2}+2)$ when $t = 2$: $5\times2^{2}+2=5\times4+2=20 + 2=22$.
Third, calculate $(10t)$ when $t = 2$: $10\times2 = 20$.
Fourth, calculate $(t^{2}+5t + 3)$ when $t = 2$: $2^{2}+5\times2+3=4 + 10+3=17$.
Then $f'(2)=9\times22+20\times17$.

Step3: Perform the multi -plications and addition

$9\times22 = 198$ and $20\times17=340$.
$f'(2)=198 + 340=538$.

Answer:

$538$