1. The problem is to find the first, second, and third derivatives of the function $y = \cos^3 x$. 2. Start with the first derivative. Use the chain rule: if $y = (u)^3$ where $u = \cos x$, then $$\frac{dy}{dx} = 3u^2 \frac{du}{dx}.$$ Since $\frac{d}{dx}(\cos x) = -\sin x$, we get $$\frac{dy}{dx} = 3 \cos^2 x (-\sin x) = -3 \cos^2 x \sin x.$$ 3. For the second derivative, differentiate $y' = -3 \cos^2 x \sin x$ using the product rule: $$\frac{d^2y}{dx^2} = -3 \left( \frac{d}{dx}(\cos^2 x) \sin x + \cos^2 x \frac{d}{dx}(\sin x) \right).$$ 4. Calculate each derivative: $$\frac{d}{dx}(\cos^2 x) = 2 \cos x (-\sin x) = -2 \cos x \sin x,$$ $$\frac{d}{dx}(\sin x) = \cos x.$$ 5. Substitute back: $$\frac{d^2y}{dx^2} = -3 \left( -2 \cos x \sin x \sin x + \cos^2 x \cos x \right) = -3 \left( -2 \cos x \sin^2 x + \cos^3 x \right).$$ Simplify: $$\frac{d^2y}{dx^2} = -3 \cos^3 x + 6 \cos x \sin^2 x.$$ 6. For the third derivative, differentiate the second derivative: $$\frac{d^3y}{dx^3} = \frac{d}{dx} \left(-3 \cos^3 x + 6 \cos x \sin^2 x \right).$$ 7. Differentiate term by term: - For $-3 \cos^3 x$, use chain rule: $$\frac{d}{dx}(-3 \cos^3 x) = -3 \cdot 3 \cos^2 x (-\sin x) = 9 \cos^2 x \sin x.$$ - For $6 \cos x \sin^2 x$, use product rule: $$\frac{d}{dx}(6 \cos x \sin^2 x) = 6 \left( -\sin x \sin^2 x + \cos x \cdot 2 \sin x \cos x \right) = 6 (-\sin x \sin^2 x + 2 \cos^2 x \sin x).$$ 8. Simplify the second term: $$6 (-\sin^3 x + 2 \cos^2 x \sin x) = -6 \sin^3 x + 12 \cos^2 x \sin x.$$ 9. Combine all parts: $$\frac{d^3y}{dx^3} = 9 \cos^2 x \sin x - 6 \sin^3 x + 12 \cos^2 x \sin x = (9 + 12) \cos^2 x \sin x - 6 \sin^3 x = 21 \cos^2 x \sin x - 6 \sin^3 x.$$ 10. Using the identity $\cos^2 x = 1 - \sin^2 x$, rewrite: $$\frac{d^3y}{dx^3} = 21 (1 - \sin^2 x) \sin x - 6 \sin^3 x = 21 \sin x - 21 \sin^3 x - 6 \sin^3 x = 21 \sin x - 27 \sin^3 x.$$ **Final answers:** $$y = \cos^3 x,$$ $$y' = -3 \cos^2 x \sin x,$$ $$y'' = -3 \cos^3 x + 6 \cos x \sin^2 x,$$ $$y''' = 21 \sin x - 27 \sin^3 x.$$