To find the derivative of the function [tex]\( S(t) \)[/tex] with respect to [tex]\( t \)[/tex], we start with the given function:
[tex]\[ S(t) = \frac{40 t^2}{t^2 + 150} \][/tex]
We can use the quotient rule to differentiate this function. The quotient rule states that if you have a function in the form [tex]\( \frac{u(t)}{v(t)} \)[/tex], where both [tex]\( u(t) \)[/tex] and [tex]\( v(t) \)[/tex] are differentiable functions of [tex]\( t \)[/tex], then the derivative is given by:
[tex]\[ \frac{d}{dt} \left( \frac{u(t)}{v(t)} \right) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} \][/tex]
In our case:
[tex]\[
u(t) = 40 t^2 \quad \text{and} \quad v(t) = t^2 + 150
\][/tex]
First, we find the derivatives [tex]\( u'(t) \)[/tex] and [tex]\( v'(t) \)[/tex]:
[tex]\[ u'(t) = \frac{d}{dt} (40 t^2) = 80 t \][/tex]
[tex]\[ v'(t) = \frac{d}{dt} (t^2 + 150) = 2 t \][/tex]
Now, we apply the quotient rule:
[tex]\[
S'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}
\][/tex]
Substitute [tex]\( u(t) \)[/tex], [tex]\( u'(t) \)[/tex], [tex]\( v(t) \)[/tex], and [tex]\( v'(t) \)[/tex] into the formula:
[tex]\[
S'(t) = \frac{(80 t)(t^2 + 150) - (40 t^2)(2 t)}{(t^2 + 150)^2}
\][/tex]
Simplify the numerator step-by-step:
[tex]\[
S'(t) = \frac{80 t (t^2 + 150) - 80 t^3}{(t^2 + 150)^2}
\][/tex]
[tex]\[
S'(t) = \frac{80 t^3 + 12000 t - 80 t^3}{(t^2 + 150)^2}
\][/tex]
[tex]\[
S'(t) = \frac{12000 t}{(t^2 + 150)^2}
\][/tex]
Combining like terms, we find that:
[tex]\[
S'(t) = \frac{-80 t^3 + 80 t (t^2 + 150)}{(t^2 + 150)^2}
\][/tex]
After simplifying, we get the final expression for the derivative:
[tex]\[
S'(t) = \frac{-80 t^3}{(t^2 + 150)^2} + \frac{80 t}{(t^2 + 150)^2}
\][/tex]
So, the derivative [tex]\( S'(t) \)[/tex] is:
[tex]\[
S'(t) = \frac{-80 t^3}{(t^2 + 150)^2} + \frac{80 t}{(t^2 + 150)}
\][/tex]
