hw13 derivatives of trigonometric\nscore: 6/7 answered: 6/7\nquestion 7\nfind $\frac{dy}{dv}$ for the given…

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, $u = 2-\cos(v)$ and $v=7 + 4\cos(v)$. First, find $u'$ and $v'$. $u'=\frac{d}{dv}(2-\cos(v))=\sin(v)$ (since $\frac{d}{dv}(2)=0$ and $\frac{d}{dv}(-\cos(v))=\sin(v)$) and $v'=\frac{d}{dv}(7 + 4\cos(v))=-4\sin(v)$ (since $\frac{d}{dv}(7)=0$ and $\frac{d}{dv}(4\cos(v))=-4\sin(v)$).

Step2: Substitute into quotient - rule formula

$\frac{dy}{dv}=\frac{\sin(v)(7 + 4\cos(v))-(2-\cos(v))(-4\sin(v))}{(7 + 4\cos(v))^{2}}$. Expand the numerator: [ \begin{align*} &\sin(v)(7 + 4\cos(v))-(2-\cos(v))(-4\sin(v))\ =&7\sin(v)+4\sin(v)\cos(v)+8\sin(v)-4\sin(v)\cos(v)\ =&15\sin(v) \end{align*} ] So, $\frac{dy}{dv}=\frac{15\sin(v)}{(7 + 4\cos(v))^{2}}$.

Answer:

$\frac{15\sin(v)}{(7 + 4\cos(v))^{2}}$