if ( f(x)=\frac{2 sin x}{1+cos x} ), then ( f^{prime}(x)= ) ( f^{prime}(2)= )

Answer:

$f'(x)=\frac{2}{1 + \cos x}$; $f'(2)=\frac{2}{1+\cos2}$

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, $u = 2\sin x$, $u'=2\cos x$, $v = 1+\cos x$, $v'=-\sin x$. [ \begin{align*} f'(x)&=\frac{(2\cos x)(1 + \cos x)-2\sin x(-\sin x)}{(1 + \cos x)^{2}}\ &=\frac{2\cos x+2\cos^{2}x + 2\sin^{2}x}{(1 + \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'(x)&=\frac{2\cos x+2(\cos^{2}x+\sin^{2}x)}{(1 + \cos x)^{2}}\ &=\frac{2\cos x + 2}{(1 + \cos x)^{2}}\ &=\frac{2(1+\cos x)}{(1 + \cos x)^{2}}\ &=\frac{2}{1+\cos x} \end{align*} ]

Step3: Evaluate $f'(x)$ at $x = 2$

Substitute $x = 2$ into $f'(x)$: $f'(2)=\frac{2}{1+\cos2}$