find the derivative of the function.\ns=(1 - t)(5 + t^{2})^{-1}\n\\frac{ds}{dt}=\\square

Explanation:

Step1: Apply product rule

The product rule states that if $s = u\cdot v$, then $\frac{ds}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$. Let $u = 1 - t$ and $v=(5 + t^{2})^{-1}$. First, find $\frac{du}{dt}$ and $\frac{dv}{dt}$. $\frac{du}{dt}=\frac{d}{dt}(1 - t)=- 1$. To find $\frac{dv}{dt}$, use the chain - rule. Let $w = 5 + t^{2}$, so $v = w^{-1}$. Then $\frac{dv}{dw}=-w^{-2}$ and $\frac{dw}{dt}=2t$. By the chain - rule $\frac{dv}{dt}=\frac{dv}{dw}\cdot\frac{dw}{dt}=- (5 + t^{2})^{-2}\cdot2t$.

Step2: Substitute into product rule

$\frac{ds}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}=(1 - t)\left(-\frac{2t}{(5 + t^{2})^{2}}\right)+(5 + t^{2})^{-1}\cdot(-1)$. Simplify the expression: [ \begin{align*} \frac{ds}{dt}&=-\frac{2t(1 - t)}{(5 + t^{2})^{2}}-\frac{1}{5 + t^{2}}\ &=-\frac{2t-2t^{2}}{(5 + t^{2})^{2}}-\frac{5 + t^{2}}{(5 + t^{2})^{2}}\ &=\frac{-2t + 2t^{2}-5 - t^{2}}{(5 + t^{2})^{2}}\ &=\frac{t^{2}-2t - 5}{(5 + t^{2})^{2}} \end{align*} ]

Answer:

$\frac{t^{2}-2t - 5}{(5 + t^{2})^{2}}$