Question

Explain 2 different ways to solve for the derivative s(θ)=200sinθcosθ

Answer 1

To find the derivative of the function s we can use the two following approaches:

0. Product rule.

,

1. Using trigonometric identities.

Product rule.

We know that the product rule states that:

`(d)/(dx)(fg)=f^(\prime)g+fg^(\prime)`

Using this rule in this case we will have:

`\begin{gathered} (ds)/(d\theta)=(d)/(d\theta)(200\sin\theta\cos\theta) \\ =200(d)/(d\theta)(\sin\theta\cos\theta) \\ =200(\cos\theta(d)/(d\theta)\sin\theta+\sin\theta(d)/(d\theta)\cos\theta) \\ =200(\cos\theta\cos\theta+\sin\theta(-\sin\theta)) \\ =200(\cos^2\theta-\sin^2\theta) \end{gathered}`

Therefore, using the product rule, we have that:

`(ds)/(d\theta)=200(\cos^2\theta-\sin^2\theta)`

Using trigonometric identities.

We can also calculate the derivative if we remember that:

`\sin2\theta=2\sin\theta\cos\theta`

Then, in this case we have:

`\begin{gathered} (ds)/(d\theta)=(d)/(d\theta)(200\sin\theta\cos\theta) \\ =100(d)/(d\theta)(2\sin\theta\cos\theta) \\ =100(d)/(d\theta)\sin2\theta \\ =100(2\cos2\theta) \\ =200\cos2\theta \end{gathered}`

Therefore, using trigonometric identities, we have that:

`(ds)/(d\theta)=200\cos2\theta`

Note: Both results are equivalent, to prove it we just need to remember that:

`\cos^2\theta-\sin^2\theta=\cos2\theta`

If we use this identity in the first result we get the second one.