Question

Find the derivatives of the function

Answer 1

Part B)

`\displaystyle h^\prime(t)=(2t^3(4-t^3))/((t^3+2)^3)`

Part C)

`h^\prime(t)=-4\pi\cot(\pi t+2)\csc^2(\pi t+2)`

Step-by-Step Explanation:

Question B)

We have:

`\displaystyle h(t)=((t^2)/(t^3+2))^2`

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:

`\displaystyle (d)/(dx)[(u(v(x))]=u^\prime(v(x))\cdot v^\prime(x)`

And the quotient rule:

`\displaystyle (d)/(dx)[(u)/(v)]=(u^\prime v-uv^\prime)/(v^2)`

Therefore, for our function h(t), we can let, by the chain rule:

`\displaystyle v(t)=(t^2)/(t^3+2)\text{ and } u(t)=t^2`

Then by the quotient rule:

`\displaystyle v^\prime(t)=(2t(t^3+2)-t^2(3t^2))/((t^3+2)^2)\text{ and } u^\prime(t)=2t`

Then, by the chain rule, our derivative, h’(t), is:

`\displaystyle h^\prime(t)=2((t^2)/(t^3+2))((2t(t^3+2)-t^2(3t^2))/((t^3+2)^2))`

Simplify:

`\begin{aligned} \displaystyle h^\prime(t)&=2((t^2)/(t^3+2))((2t(t^3+2)-t^2(3t^2))/((t^3+2)^2))\\ \\ &=(2t^2)/(t^3+2)((2t^4+4t-3t^4)/((t^3+2)^2))\\ \\ &= (2t^2)/(t^3+2)((-t^4+4t)/((t^3+2)^2)) \\ \\ &=(2t^3(4-t^3))/((t^3+2)^3) \end{aligned}`

Part C)

We have:

`h(t)=2\cot^2(\pi t+2)`

Again, we will utilize the chain rule. This time, we will let:

`u(t)=x^2\text{ and } v(t)=\cot(\pi t+2)`

Then differentiating gives (on the right, we will apply the chain rule a second time):

`u^\prime(t)=2t\text{ and } v^\prime(t)=-\pi\csc^2(\pi t+2)`

To differentiate v(t), as mentioned, we need to apply the chain rule. We have:

`v(t)=\cot(\pi t+2)`

We will let:

`u_2(t)=\cot(t)\text{ and } v_2(t)=\pi t+2`

Therefore:

`u_2^\prime(t)=-\csc^2(t)\text{ and } v_2^\prime(t)=\pi`

So:

`v^\prime(t)=-\csc^2(\pi t+2)(\pi)`

And by simplification:

`v^\prime(t)=-\pi\csc^2(\pi t+2)`

Therefore, it follows that:

`h^\prime(t)=2[2(\cot(\pi t+2))(-\pi \csc^2(\pi t+2))]`

So:

`h^\prime(t)=-4\pi\cot(\pi t+2)\csc^2(\pi t+2)`