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63. $y = \\frac{\\theta + 5}{\\theta\\cos\\theta}$

Question

  1. $y = \frac{\theta + 5}{\theta\cos\theta}$

Explanation:

Step1: Apply the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \theta+5$, so $u^\prime=1$, and $v=\theta\cos\theta$.
First, find $v^\prime$ using the product - rule. The product - rule states that if $v = ab$ (where $a = \theta$ and $b=\cos\theta$), then $v^\prime=a^\prime b+ab^\prime$. Since $a^\prime = 1$ and $b^\prime=-\sin\theta$, we have $v^\prime=\cos\theta-\theta\sin\theta$.

Step2: Substitute $u$, $u^\prime$, $v$, and $v^\prime$ into the quotient - rule

$y^\prime=\frac{1\times(\theta\cos\theta)-(\theta + 5)(\cos\theta-\theta\sin\theta)}{(\theta\cos\theta)^{2}}$.
Expand the numerator:
\[

$$\begin{align*} &(\theta\cos\theta)-(\theta\cos\theta-\theta^{2}\sin\theta + 5\cos\theta-5\theta\sin\theta)\\ =&\theta\cos\theta-\theta\cos\theta+\theta^{2}\sin\theta - 5\cos\theta+5\theta\sin\theta\\ =&\theta^{2}\sin\theta+5\theta\sin\theta - 5\cos\theta \end{align*}$$

\]
So, $y^\prime=\frac{\theta^{2}\sin\theta + 5\theta\sin\theta-5\cos\theta}{\theta^{2}\cos^{2}\theta}$.

Answer:

$y^\prime=\frac{\theta^{2}\sin\theta + 5\theta\sin\theta-5\cos\theta}{\theta^{2}\cos^{2}\theta}$