Answer:
Part B)
[tex]\displaystyle h^\prime(t)=\frac{2t^3(4-t^3)}{(t^3+2)^3}[/tex]
Part C)
[tex]h^\prime(t)=-4\pi\cot(\pi t+2)\csc^2(\pi t+2)[/tex]
Step-by-Step Explanation:
Question B)
We have:
[tex]\displaystyle h(t)=(\frac{t^2}{t^3+2})^2[/tex]
And we want to find the derivative, h‘(t).
This will require the chain rule and the quotient rule. Remember that the chain rule states:
[tex]\displaystyle \frac{d}{dx}[(u(v(x))]=u^\prime(v(x))\cdot v^\prime(x)[/tex]
And the quotient rule:
[tex]\displaystyle \frac{d}{dx}[\frac{u}{v}]=\frac{u^\prime v-uv^\prime}{v^2}[/tex]
Therefore, for our function h(t), we can let, by the chain rule:
[tex]\displaystyle v(t)=\frac{t^2}{t^3+2}\text{ and } u(t)=t^2[/tex]
Then by the quotient rule:
[tex]\displaystyle v^\prime(t)=\frac{2t(t^3+2)-t^2(3t^2)}{(t^3+2)^2}\text{ and } u^\prime(t)=2t[/tex]
Then, by the chain rule, our derivative, h’(t), is:
[tex]\displaystyle h^\prime(t)=2(\frac{t^2}{t^3+2})(\frac{2t(t^3+2)-t^2(3t^2)}{(t^3+2)^2})[/tex]
Simplify:
[tex]\begin{aligned} \displaystyle h^\prime(t)&=2(\frac{t^2}{t^3+2})(\frac{2t(t^3+2)-t^2(3t^2)}{(t^3+2)^2})\\ \\ &=\frac{2t^2}{t^3+2}(\frac{2t^4+4t-3t^4}{(t^3+2)^2})\\ \\ &= \frac{2t^2}{t^3+2}(\frac{-t^4+4t}{(t^3+2)^2}) \\ \\ &=\frac{2t^3(4-t^3)}{(t^3+2)^3} \end{aligned}[/tex]
Part C)
We have:
[tex]h(t)=2\cot^2(\pi t+2)[/tex]
Again, we will utilize the chain rule. This time, we will let:
[tex]u(t)=x^2\text{ and } v(t)=\cot(\pi t+2)[/tex]
Then differentiating gives (on the right, we will apply the chain rule a second time):
[tex]u^\prime(t)=2t\text{ and } v^\prime(t)=-\pi\csc^2(\pi t+2)[/tex]
To differentiate v(t), as mentioned, we need to apply the chain rule. We have:
[tex]v(t)=\cot(\pi t+2)[/tex]
We will let:
[tex]u_2(t)=\cot(t)\text{ and } v_2(t)=\pi t+2[/tex]
Therefore:
[tex]u_2^\prime(t)=-\csc^2(t)\text{ and } v_2^\prime(t)=\pi[/tex]
So:
[tex]v^\prime(t)=-\csc^2(\pi t+2)(\pi)[/tex]
And by simplification:
[tex]v^\prime(t)=-\pi\csc^2(\pi t+2)[/tex]
Therefore, it follows that:
[tex]h^\prime(t)=2[2(\cot(\pi t+2))(-\pi \csc^2(\pi t+2))][/tex]
So:
[tex]h^\prime(t)=-4\pi\cot(\pi t+2)\csc^2(\pi t+2)[/tex]
