1. Differentiate $y = 2x^3 + 5x^2 - 4x + 9$. The derivative of a polynomial is found by applying the power rule: $\frac{d}{dx} x^n = n x^{n-1}$. Step 1: Differentiate each term separately: $$\frac{d}{dx}(2x^3) = 2 \times 3 x^{3-1} = 6x^2$$ $$\frac{d}{dx}(5x^2) = 5 \times 2 x^{2-1} = 10x$$ $$\frac{d}{dx}(-4x) = -4$$ $$\frac{d}{dx}(9) = 0$$ Step 2: Combine the results: $$\frac{dy}{dx} = 6x^2 + 10x - 4$$ This is the derivative of the first function. --- 2. Differentiate $y = (2 - 3x^2)^{-4}$. Use the chain rule: $\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)$. Step 1: Let $u = 2 - 3x^2$, then $y = u^{-4}$. Step 2: Differentiate outer function: $$\frac{d}{du} u^{-4} = -4 u^{-5}$$ Step 3: Differentiate inner function: $$\frac{d}{dx} (2 - 3x^2) = -6x$$ Step 4: Apply chain rule: $$\frac{dy}{dx} = -4 u^{-5} \times (-6x) = 24x (2 - 3x^2)^{-5}$$ --- 3. Differentiate $y = (1 - 2x)(3x + 2)^4$. Use the product rule: $\frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$. Step 1: Let $f = 1 - 2x$, $g = (3x + 2)^4$. Step 2: Differentiate $f$: $$f' = -2$$ Step 3: Differentiate $g$ using chain rule: $$g' = 4(3x + 2)^3 \times 3 = 12(3x + 2)^3$$ Step 4: Apply product rule: $$\frac{dy}{dx} = (-2)(3x + 2)^4 + (1 - 2x) \times 12(3x + 2)^3$$ Step 5: Factor out $(3x + 2)^3$: $$\frac{dy}{dx} = (3x + 2)^3 \left[-2(3x + 2) + 12(1 - 2x)\right]$$ Step 6: Simplify inside brackets: $$-2(3x + 2) + 12(1 - 2x) = -6x - 4 + 12 - 24x = (-6x - 24x) + (-4 + 12) = -30x + 8$$ Final derivative: $$\frac{dy}{dx} = (3x + 2)^3 (-30x + 8)$$ --- 4. Differentiate $y = \frac{5x}{2x + 1}$. Use the quotient rule: $\frac{d}{dx} \left( \frac{f}{g} \right) = \frac{f'g - fg'}{g^2}$. Step 1: Let $f = 5x$, $g = 2x + 1$. Step 2: Differentiate $f$ and $g$: $$f' = 5$$ $$g' = 2$$ Step 3: Apply quotient rule: $$\frac{dy}{dx} = \frac{5(2x + 1) - 5x(2)}{(2x + 1)^2}$$ Step 4: Simplify numerator: $$5(2x + 1) - 10x = 10x + 5 - 10x = 5$$ Step 5: Final derivative: $$\frac{dy}{dx} = \frac{5}{(2x + 1)^2}$$ --- 5. Differentiate $y = \cos^3 x$. Rewrite as $y = (\cos x)^3$ and use chain rule. Step 1: Let $u = \cos x$, then $y = u^3$. Step 2: Differentiate outer function: $$\frac{d}{du} u^3 = 3u^2$$ Step 3: Differentiate inner function: $$\frac{d}{dx} \cos x = -\sin x$$ Step 4: Apply chain rule: $$\frac{dy}{dx} = 3(\cos x)^2 \times (-\sin x) = -3 \cos^2 x \sin x$$