Question

In Exercises \(37-42,\) find \(f^{\prime}(x)\) and state the domain of \(f^{\prime}\) $$f(x)=\ln (2-\cos x)$$

Short answer

The derivative of the function \( f(x) = \ln(2- \cos(x)) \) is \( f^{'}(x)=\frac{\sin(x)}{2-\cos(x)} \), and the domain of \( f^{'}(x) \) is all real numbers.

Step-by-step solution

Identify the Inner Function

There's a composite function defined as \( f(x) = \ln(2- \cos(x)) \). The outer function is \( \ln(u) \) and the inner function is \( u=2- \cos(x) \) where \( u \) is a function of \( x \).

Apply the Chain Rule for differentiation

The chain rule for finding the derivative of a composite function is \( (f \circ g)^{'}(x) = f^{'}(g(x)) \cdot g^{'}(x) \). The derivative of the outer function \( \ln(u) \) is \( f^{'}(u)=\frac{1}{u} \). Now, plug in \( u=2-\cos(x) \) into \( f^{'}(u) \) to get \( f^{'}(2-\cos(x))=\frac{1}{2-\cos(x)} \). The derivative of the inner function \( g(x)=2-\cos(x) \) is \( g^{'}(x)=\sin(x) \). Finally, multiply \( f^{'}(2-\cos(x)) \) by \( g^{'}(x) \) to get \( f^{'}(x)=\frac{\sin(x)}{2-\cos(x)} \).

Find the domain of the derivative function

The domain of \( f^{'}(x) \) is all \( x \) such that \( 2-\cos(x) \neq 0 \). The cosine function oscillates between -1 and 1, inclusive. Therefore, \( 2-\cos(x) \) varies from 1 (when \( \cos(x) = 1 \)) to 3 (when \( \cos(x) = -1 \)). Hence, the value of \( 2- \cos(x) \) will never be zero for any real value of \( x \). Therefore, the domain of the derivative function is all real numbers.