# Math Help - Product and Quotient Rule

1. ## Product and Quotient Rule

In B.C, the sales of specialized high-endurance, out dollar gear in millions of dollars is projected by

s(t)= 4t/t^2+2
0<t<10

where t is measured in years with t=0 corresponding to the beginning of 2000. How fast was the level of sales increasing at the beginning of 2005?

2. Originally Posted by lemontea
In B.C, the sales of specialized high-endurance, out dollar gear in millions of dollars is projected by

s(t)= 4t/t^2+2
0<t<10

where t is measured in years with t=0 corresponding to the beginning of 2000. How fast was the level of sales increasing at the beginning of 2005?
Do you mean : $s(t)=\frac{4t}{t^2+2}$?

3. Originally Posted by Chris L T521
Do you mean : $s(t)=\frac{4t}{t^2+2}$?

If this is the case, then find $s'(5)$

Using Quotient Rule:

$s'(t)=\frac{(t^2+2)(4)-(4t)(2t)}{(t^2+2)^2}\implies s'(t)=\frac{-4t^2+8}{(t^2+2)^2}$

Thus, $\color{red}\boxed{s'(5)=-\frac{92}{27^2}\approx -.1262}$

Using Product Rule:

$s(t)=(4t)(t^2+2)^{-1}$

$\therefore s'(t)=4(t^2+2)^{-1}-(4t)(t^2+2)^{-2}\cdot (2t)$
$\implies s'(t)=\frac{4}{t^2+2}-\frac{8t^2}{(t^2+2)^2}$
$\implies s'(t)=\frac{4t^2+8-8t^2}{(t^2+2)^2}=\frac{-4t^2+8}{(t^2+2)^2}$

Thus, $\color{red}\boxed{s'(5)=-\frac{92}{27^2}\approx -.1262}$