Questions: If f(x) = 2 sin x / (3 + cos x), then f'(x) = f'(5) =

Solution Steps

To find the derivative of the function \( f(x) = \frac{2 \sin x}{3 + \cos x} \), we will use the quotient rule for differentiation. The quotient rule states that if you have a function \( g(x) = \frac{u(x)}{v(x)} \), then its derivative is given by \( g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). Here, \( u(x) = 2 \sin x \) and \( v(x) = 3 + \cos x \). After finding the derivative \( f'(x) \), we will evaluate it at \( x = 5 \).

Given the function \( f(x) = \frac{2 \sin x}{3 + \cos x} \), we need to find its derivative using the quotient rule. The quotient rule states that if \( g(x) = \frac{u(x)}{v(x)} \), then \( g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).

For \( f(x) \), let:

• \( u(x) = 2 \sin x \) with \( u'(x) = 2 \cos x \)
• \( v(x) = 3 + \cos x \) with \( v'(x) = -\sin x \)

Applying the quotient rule: \[ f'(x) = \frac{(2 \cos x)(3 + \cos x) - (2 \sin x)(-\sin x)}{(3 + \cos x)^2} \]

Simplify the expression for \( f'(x) \): \[ f'(x) = \frac{2 \cos x (3 + \cos x) + 2 \sin^2 x}{(3 + \cos x)^2} \]

This simplifies to: \[ f'(x) = \frac{2 \cos x + 2 \cos^2 x + 2 \sin^2 x}{(3 + \cos x)^2} \]

Using the identity \( \sin^2 x + \cos^2 x = 1 \), we further simplify: \[ f'(x) = \frac{2 \cos x + 2}{(3 + \cos x)^2} \]

Substitute \( x = 5 \) into the derivative: \[ f'(5) = \frac{2 \cos(5) + 2}{(3 + \cos(5))^2} \]

Final Answer