if ( f(x)=\frac{3 sin x}{2+cos x} ), then ( f^{prime}(x)= ) ( f^{prime}(5)= )

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 = 3\sin x), (u^\prime=3\cos x), (v = 2+\cos x), (v^\prime=-\sin x). [ \begin{align*} f^\prime(x)&=\frac{(3\cos x)(2 + \cos x)-3\sin x(-\sin x)}{(2+\cos x)^{2}}\ &=\frac{6\cos x+3\cos^{2}x + 3\sin^{2}x}{(2+\cos x)^{2}} \end{align*} ]

Step2: Use the trigonometric identity (\sin^{2}x+\cos^{2}x = 1)

Substitute (\sin^{2}x+\cos^{2}x = 1) into the numerator. [ \begin{align*} f^\prime(x)&=\frac{6\cos x+3(\cos^{2}x+\sin^{2}x)}{(2+\cos x)^{2}}\ &=\frac{6\cos x + 3\times1}{(2+\cos x)^{2}}\ &=\frac{6\cos x+3}{(2+\cos x)^{2}} \end{align*} ]

Step3: Calculate (f^\prime(5))

Substitute (x = 5) into (f^\prime(x)). [ f^\prime(5)=\frac{6\cos(5)+3}{(2+\cos(5))^{2}} ] Using a calculator (in radian mode), (\cos(5)\approx0.283662) [ \begin{align*} f^\prime(5)&=\frac{6\times0.283662+3}{(2 + 0.283662)^{2}}\ &=\frac{1.701972+3}{(2.283662)^{2}}\ &=\frac{4.701972}{5.215077}\ &\approx0.9016 \end{align*} ]

Answer:

(f^\prime(x)=\frac{6\cos x + 3}{(2+\cos x)^{2}}), (f^\prime(5)\approx0.9016)