Question

if $f(t)=(t^{2}+4t + 5)(3t^{-3}+5t^{-2})$, find $f(t)$. $f(t)=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=t^{2}+4t + 5$ and $v = 3t^{-3}+5t^{-2}$. First, find $u'$ and $v'$.
$u'=\frac{d}{dt}(t^{2}+4t + 5)=2t + 4$
$v'=\frac{d}{dt}(3t^{-3}+5t^{-2})=-9t^{-4}-10t^{-3}$

Step2: Substitute into product - rule

$f'(t)=u'v+uv'=(2t + 4)(3t^{-3}+5t^{-2})+(t^{2}+4t + 5)(-9t^{-4}-10t^{-3})$
Expand the two products:
$(2t + 4)(3t^{-3}+5t^{-2})=2t\cdot3t^{-3}+2t\cdot5t^{-2}+4\cdot3t^{-3}+4\cdot5t^{-2}=6t^{-2}+10t^{-1}+12t^{-3}+20t^{-2}=26t^{-2}+10t^{-1}+12t^{-3}$
$(t^{2}+4t + 5)(-9t^{-4}-10t^{-3})=-9t^{-2}-10t^{-1}-36t^{-3}-40t^{-2}-45t^{-4}-50t^{-3}=-49t^{-2}-10t^{-1}-86t^{-3}-45t^{-4}$

Step3: Combine like - terms

$f'(t)=(26t^{-2}+10t^{-1}+12t^{-3})+(-49t^{-2}-10t^{-1}-86t^{-3}-45t^{-4})$
$=-23t^{-2}-74t^{-3}-45t^{-4}$
$=-\frac{23}{t^{2}}-\frac{74}{t^{3}}-\frac{45}{t^{4}}$

Answer:

$-\frac{23}{t^{2}}-\frac{74}{t^{3}}-\frac{45}{t^{4}}$