The given function is:

$g(x) = x^{\cos(x)}$

To find the derivative $g'(x)$ of this function, we can use logarithmic differentiation, which is especially useful for functions where the exponent is also a function of $x$.

Step-by-step solution:

1. Take the natural logarithm of both sides: $\ln(g(x)) = \ln(x^{\cos(x)})$

2. Simplify using logarithm properties: $\ln(g(x)) = \cos(x) \ln(x)$

3. Differentiate both sides with respect to $x$: Use the chain rule on the left: $\frac{g'(x)}{g(x)} = \cos(x) \cdot \frac{1}{x} + \ln(x) \cdot (-\sin(x))$

   Simplifying this gives: $\frac{g'(x)}{g(x)} = \frac{\cos(x)}{x} - \sin(x) \ln(x)$

4. Solve for $g'(x)$: Multiply both sides by $g(x) = x^{\cos(x)}$: $g'(x) = x^{\cos(x)} \left( \frac{\cos(x)}{x} - \sin(x) \ln(x) \right)$

Final derivative:

$g'(x) = x^{\cos(x)} \left( \frac{\cos(x)}{x} - \sin(x) \ln(x) \right)$

Would you like more details or have any specific questions on this?

Here are 5 related questions to deepen understanding:

1. How do you apply logarithmic differentiation in general?
2. What is the domain of the function $g(x) = x^{\cos(x)}$?
3. How would you find the critical points of $g(x)$?
4. Can $g(x)$ have any asymptotes?
5. How does $g(x)$ behave as $x \to 0^+$ and $x \to \infty$?

Tip: Logarithmic differentiation is especially useful when dealing with expressions like $f(x)^{g(x)}$, where both the base and exponent are functions of $x$.