1. **State the problem:** We need to find the derivative of the function $$y = \sin^6 \theta - \cos^5 \theta$$ with respect to $$\theta$$. 2. **Recall the formula and rules:** - The chain rule states that $$\frac{d}{d\theta}[f(g(\theta))] = f'(g(\theta)) \cdot g'(\theta)$$. - The derivative of $$\sin \theta$$ is $$\cos \theta$$. - The derivative of $$\cos \theta$$ is $$-\sin \theta$$. 3. **Apply the chain rule to each term:** - For $$\sin^6 \theta$$, treat it as $$[\sin \theta]^6$$. - For $$\cos^5 \theta$$, treat it as $$[\cos \theta]^5$$. 4. **Differentiate the first term:** $$\frac{d}{d\theta} \sin^6 \theta = 6 \sin^5 \theta \cdot \cos \theta$$ 5. **Differentiate the second term:** $$\frac{d}{d\theta} \cos^5 \theta = 5 \cos^4 \theta \cdot (-\sin \theta) = -5 \cos^4 \theta \sin \theta$$ 6. **Combine the derivatives:** $$\frac{dy}{d\theta} = 6 \sin^5 \theta \cos \theta - (-5 \cos^4 \theta \sin \theta) = 6 \sin^5 \theta \cos \theta + 5 \cos^4 \theta \sin \theta$$ 7. **Factor common terms:** $$\frac{dy}{d\theta} = \sin \theta \cos \theta (6 \sin^4 \theta + 5 \cos^3 \theta)$$ **Final answer:** $$\boxed{\frac{dy}{d\theta} = \sin \theta \cos \theta (6 \sin^4 \theta + 5 \cos^3 \theta)}$$