Question

hw12 derivatives of trigonometric functions (target score: 6/7 answered: 6/7 question 7 find $\frac{dy}{dv}$ for the given function. $y = \frac{5 - sin(v)}{4 + 9sin(v)}$ $\frac{dy}{dv}=$ question help: message instructor

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dv}=\frac{u'v - uv'}{v^{2}}$. Here, $u = 5-\sin(v)$ and $v = 4 + 9\sin(v)$. First, find $u'$ and $v'$.
$u'=\frac{d}{dv}(5-\sin(v))=0-\cos(v)=-\cos(v)$
$v'=\frac{d}{dv}(4 + 9\sin(v))=0+9\cos(v)=9\cos(v)$

Step2: Substitute into quotient - rule formula

$\frac{dy}{dv}=\frac{(-\cos(v))(4 + 9\sin(v))-(5-\sin(v))(9\cos(v))}{(4 + 9\sin(v))^{2}}$
Expand the numerator:
\[

$$\begin{align*} &(-\cos(v))(4 + 9\sin(v))-(5-\sin(v))(9\cos(v))\\ =&-4\cos(v)-9\sin(v)\cos(v)-(45\cos(v)-9\sin(v)\cos(v))\\ =&-4\cos(v)-9\sin(v)\cos(v)-45\cos(v)+9\sin(v)\cos(v)\\ =&-49\cos(v) \end{align*}$$

\]
So, $\frac{dy}{dv}=\frac{-49\cos(v)}{(4 + 9\sin(v))^{2}}$

Answer:

$\frac{-49\cos(v)}{(4 + 9\sin(v))^{2}}$