1. **State the problem:** Find the derivative of the function $$f(x) = (\cos x)^{\sin x}$$ for $$x > 0$$. 2. **Recall the formula:** For a function of the form $$y = u(x)^{v(x)}$$, the derivative is given by $$ \frac{dy}{dx} = u(x)^{v(x)} \left( v'(x) \ln u(x) + v(x) \frac{u'(x)}{u(x)} \right) $$ where $$u(x) = \cos x$$ and $$v(x) = \sin x$$. 3. **Calculate derivatives:** - $$u'(x) = -\sin x$$ - $$v'(x) = \cos x$$ 4. **Substitute into the formula:** $$ \frac{dy}{dx} = (\cos x)^{\sin x} \left( \cos x \ln(\cos x) + \sin x \frac{-\sin x}{\cos x} \right) $$ 5. **Simplify the expression inside the parentheses:** $$ \cos x \ln(\cos x) - \frac{\sin^2 x}{\cos x} $$ 6. **Final derivative:** $$ \frac{dy}{dx} = (\cos x)^{\sin x} \left( \cos x \ln(\cos x) - \frac{\sin^2 x}{\cos x} \right) $$ This matches option (b). **Answer:** (b) $$(\cos x)^{\sin x} \left( -\frac{\sin^2 x}{\cos x} + \cos x \ln(\cos x) \right)$$