Question

Find derivative problem
Find B’(6)

Answer 1

`B^\prime(6) \approx -28.17`

Explanation:

We have:

`\displaystyle B(t)=24.6\sin((\pi t)/(10))(8-t)`

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

`\displaystyle B^\prime(t)=(d)/(dt)[24.6\sin((\pi t)/(10))(8-t)]`

We can move the constant outside:

`\displaystyle B^\prime(t)=24.6(d)/(dt)[\sin((\pi t)/(10))(8-t)]`

Now, we will utilize the product rule. The product rule is:

`(uv)^\prime=u^\prime v+u v^\prime`

We will let:

`\displaystyle u=\sin((\pi t)/(10))\text{ and } \\ \\ v=8-t`

Then:

`\displaystyle u^\prime=(\pi)/(10)\cos((\pi t)/(10))\text{ and } \\ \\ v^\prime= -1`

(The derivative of u was determined using the chain rule.)

Then it follows that:

`\displaystyle \begin{aligned} B^\prime(t)&=24.6(d)/(dt)[\sin((\pi t)/(10))(8-t)] \\ \\ &=24.6[((\pi)/(10)\cos((\pi t)/(10)))(8-t) - \sin((\pi t)/(10))] \end{aligned}`

Therefore:

`\displaystyle B^\prime(6) =24.6[((\pi)/(10)\cos((\pi (6))/(10)))(8-(6))- \sin((\pi (6))/(10))]`

By simplification:

`\displaystyle B^\prime(6)=24.6 [(\pi)/(10)\cos((3\pi)/(5))(2)-\sin((3\pi)/(5))] \approx -28.17`

So, the slope of the tangent line to the point (6, B(6)) is -28.17.