homework4: problem 6\n(1 point)\ndifferentiate ( y = \frac { 6 + sin x } { x + cos x } ).\n( y ^ { prime } =…

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}})