Question

hw 10 - product and quotient rules section 2.5: problem 10 (1 point) differentiate the equation, $y = \frac{t^{3}+t}{t^{4}+1}$ $y = $

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = t^{3}+t$, $v=t^{4}+1$.

Step2: Find $u^\prime$

Differentiate $u = t^{3}+t$ with respect to $t$. Using the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we get $u^\prime=\frac{d}{dt}(t^{3}+t)=3t^{2}+1$.

Step3: Find $v^\prime$

Differentiate $v=t^{4}+1$ with respect to $t$. Using the power - rule, we get $v^\prime=\frac{d}{dt}(t^{4}+1)=4t^{3}$.

Step4: Apply quotient - rule

Substitute $u$, $u^\prime$, $v$, and $v^\prime$ into the quotient - rule formula:
\[

$$\begin{align*} y^\prime&=\frac{(3t^{2}+1)(t^{4}+1)-(t^{3}+t)\times4t^{3}}{(t^{4}+1)^{2}}\\ &=\frac{3t^{6}+3t^{2}+t^{4}+1-(4t^{6}+4t^{4})}{(t^{4}+1)^{2}}\\ &=\frac{3t^{6}+3t^{2}+t^{4}+1 - 4t^{6}-4t^{4}}{(t^{4}+1)^{2}}\\ &=\frac{-t^{6}-3t^{4}+3t^{2}+1}{(t^{4}+1)^{2}} \end{align*}$$

\]

Answer:

$\frac{-t^{6}-3t^{4}+3t^{2}+1}{(t^{4}+1)^{2}}$