Question

consider the following.
g(θ)=θ - cos(θ)
h(θ)=sin(θ)
find g(θ) and h(θ).
g(θ)=1 + sin(θ)
good job.
h(θ)=cos(θ)
nice job.
differentiate.
f(θ)=(θ - cos(θ))sin(θ)

Explanation:

Step1: Identify the product - rule

The product - rule states that if $y = u(\theta)v(\theta)$, then $y'=u'(\theta)v(\theta)+u(\theta)v'(\theta)$. Here, $u(\theta)=\theta - \cos(\theta)$ and $v(\theta)=\sin(\theta)$.

Step2: Use the product - rule

We know that $u'(\theta)=1 + \sin(\theta)$ and $v'(\theta)=\cos(\theta)$. Then $f'(\theta)=u'(\theta)v(\theta)+u(\theta)v'(\theta)$.
Substitute $u(\theta),u'(\theta),v(\theta),v'(\theta)$ into the formula:
\[

$$\begin{align*} f'(\theta)&=(1 + \sin(\theta))\sin(\theta)+(\theta - \cos(\theta))\cos(\theta)\\ &=\sin(\theta)+\sin^{2}(\theta)+\theta\cos(\theta)-\cos^{2}(\theta)\\ &=\sin(\theta)+\theta\cos(\theta)-(\cos^{2}(\theta)-\sin^{2}(\theta))\\ &=\sin(\theta)+\theta\cos(\theta)-\cos(2\theta) \end{align*}$$

\]

Answer:

$f'(\theta)=\sin(\theta)+\theta\cos(\theta)-\cos(2\theta)$