Question

1. - / 1 points if $g(\theta)=\frac{sin(\theta)}{\theta}$, find $g(\theta)$ and $g(\theta)$. $g(\theta)=\thetacos(\theta)+sin(\theta)$ $g(\theta)=i$ resources read it

Explanation:

Step1: Use quotient - rule for first - derivative

The quotient rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = \sin(\theta)$ and $v=\theta$. So $u'=\cos(\theta)$ and $v' = 1$. Then $g'(\theta)=\frac{\cos(\theta)\cdot\theta-\sin(\theta)\cdot1}{\theta^{2}}=\frac{\theta\cos(\theta)-\sin(\theta)}{\theta^{2}}$. (The given $g'(\theta)$ in the problem is incorrect).

Step2: Use quotient - rule for second - derivative

Let $u=\theta\cos(\theta)-\sin(\theta)$ and $v = \theta^{2}$. Then $u'=\cos(\theta)-\theta\sin(\theta)-\cos(\theta)=-\theta\sin(\theta)$ and $v' = 2\theta$.
By the quotient rule $g''(\theta)=\frac{(-\theta\sin(\theta))\cdot\theta^{2}-(\theta\cos(\theta)-\sin(\theta))\cdot2\theta}{\theta^{4}}=\frac{-\theta^{3}\sin(\theta)- 2\theta^{2}\cos(\theta)+2\theta\sin(\theta)}{\theta^{4}}=\frac{-\theta^{2}\sin(\theta)-2\theta\cos(\theta) + 2\sin(\theta)}{\theta^{3}}$.

Answer:

$g'(\theta)=\frac{\theta\cos(\theta)-\sin(\theta)}{\theta^{2}}$; $g''(\theta)=\frac{2\sin(\theta)-\theta^{2}\sin(\theta)-2\theta\cos(\theta)}{\theta^{3}}$