Question

hw 10 - product and quotient rules section 2.5: problem 5 (1 point) if $f(t)=(t^{2}+3t + 7)(2t^{-2}+5t^{-3})$, find $f(t)$. answer:

Explanation:

Step1: Recall product - rule

The product - rule states that if $y = u\cdot v$, where $u$ and $v$ are functions of $t$, then $y^\prime=u^\prime v + uv^\prime$. Let $u=t^{2}+3t + 7$ and $v = 2t^{-2}+5t^{-3}$.

Step2: Find $u^\prime$ and $v^\prime$

Differentiate $u$ with respect to $t$: $u^\prime=\frac{d}{dt}(t^{2}+3t + 7)=2t + 3$. Differentiate $v$ with respect to $t$: $v^\prime=\frac{d}{dt}(2t^{-2}+5t^{-3})=2\times(-2)t^{-3}+5\times(-3)t^{-4}=-4t^{-3}-15t^{-4}$.

Step3: Apply the product - rule

$f^\prime(t)=u^\prime v+uv^\prime=(2t + 3)(2t^{-2}+5t^{-3})+(t^{2}+3t + 7)(-4t^{-3}-15t^{-4})$.
Expand the first product:
\[

$$\begin{align*} (2t + 3)(2t^{-2}+5t^{-3})&=2t\times2t^{-2}+2t\times5t^{-3}+3\times2t^{-2}+3\times5t^{-3}\\ &=4t^{-1}+10t^{-2}+6t^{-2}+15t^{-3}\\ &=4t^{-1}+16t^{-2}+15t^{-3} \end{align*}$$

\]
Expand the second product:
\[

$$\begin{align*} (t^{2}+3t + 7)(-4t^{-3}-15t^{-4})&=t^{2}\times(-4t^{-3})+t^{2}\times(-15t^{-4})+3t\times(-4t^{-3})+3t\times(-15t^{-4})+7\times(-4t^{-3})+7\times(-15t^{-4})\\ &=-4t^{-1}-15t^{-2}-12t^{-2}-45t^{-3}-28t^{-3}-105t^{-4}\\ &=-4t^{-1}-27t^{-2}-73t^{-3}-105t^{-4} \end{align*}$$

\]

Step4: Combine the two expanded products

\[

$$\begin{align*} f^\prime(t)&=(4t^{-1}+16t^{-2}+15t^{-3})+(-4t^{-1}-27t^{-2}-73t^{-3}-105t^{-4})\\ &=(4t^{-1}-4t^{-1})+(16t^{-2}-27t^{-2})+(15t^{-3}-73t^{-3})-105t^{-4}\\ &=- 11t^{-2}-58t^{-3}-105t^{-4} \end{align*}$$

\]

Answer:

$-11t^{-2}-58t^{-3}-105t^{-4}$