homework4: problem 6
(1 point)
differentiate ( y = \frac { 6 + sin x } { x + cos x } ).
( y ^ { prime } = )
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# Explanation:
## Step1: Apply the quotient rule
The quotient rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 6+\sin x$, so $u'=\cos x$; $v=x + \cos x$, so $v'=1-\sin x$.
## Step2: Substitute $u$, $u'$, $v$, $v'$ into the quotient rule formula
$$
\begin{align*}
y'&=\frac{\cos x(x + \cos x)-(6+\sin x)(1 - \sin x)}{(x+\cos x)^{2}}\
&=\frac{x\cos x+\cos^{2}x-(6 - 6\sin x+\sin x-\sin^{2}x)}{(x+\cos x)^{2}}\
&=\frac{x\cos x+\cos^{2}x - 6 + 5\sin x+\sin^{2}x}{(x+\cos x)^{2}}
\end{align*}
$$
Since $\sin^{2}x+\cos^{2}x = 1$, we have:
$$
\begin{align*}
y'&=\frac{x\cos x+1 - 6 + 5\sin x}{(x+\cos x)^{2}}\
&=\frac{x\cos x+5\sin x - 5}{(x+\cos x)^{2}}
\end{align*}
$$

# Answer:
$\frac{x\cos x + 5\sin x-5}{(x + \cos x)^{2}}$

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Question

Explanation:

Step1: Apply the quotient rule

Step2: Substitute \(u\), \(u'\), \(v\), \(v'\) into the quotient rule formula

Answer: