if f(t)=(t² + 3t + 8)(5t⁻² + 4t⁻³), find f(t). answer: -10t⁵ + 8t⁻² - 126t⁴ - 116t³ + 156⁻¹

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u(t)v(t)$, then $y^\prime=u^\prime(t)v(t)+u(t)v^\prime(t)$. Let $u(t)=t^{2}+3t + 8$ and $v(t)=5t^{-2}+4t^{-3}$. First, find $u^\prime(t)$ and $v^\prime(t)$. $u^\prime(t)=\frac{d}{dt}(t^{2}+3t + 8)=2t + 3$ $v^\prime(t)=\frac{d}{dt}(5t^{-2}+4t^{-3})=-10t^{-3}-12t^{-4}$

Step2: Calculate $u^\prime(t)v(t)$

$u^\prime(t)v(t)=(2t + 3)(5t^{-2}+4t^{-3})=10t^{-1}+8t^{-2}+15t^{-2}+12t^{-3}=10t^{-1}+23t^{-2}+12t^{-3}$

Step3: Calculate $u(t)v^\prime(t)$

$u(t)v^\prime(t)=(t^{2}+3t + 8)(-10t^{-3}-12t^{-4})=-10t^{-1}-12t^{-2}-30t^{-2}-36t^{-3}-80t^{-3}-96t^{-4}=-10t^{-1}-42t^{-2}-116t^{-3}-96t^{-4}$

Step4: Find $f^\prime(t)$

$f^\prime(t)=u^\prime(t)v(t)+u(t)v^\prime(t)=(10t^{-1}+23t^{-2}+12t^{-3})+(-10t^{-1}-42t^{-2}-116t^{-3}-96t^{-4})=- 19t^{-2}-104t^{-3}-96t^{-4}$

It seems the given answer in the picture is incorrect. The correct derivative $f^\prime(t)=-19t^{-2}-104t^{-3}-96t^{-4}$