1. Problem: Find the derivative of $y = x^{12}(1 + x^2)$ using the product rule. 2. Formula: The product rule states that if $y = uv$, then $$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$$ where $u$ and $v$ are functions of $x$. 3. Given: Let $u = x^{12}$ and $v = 1 + x^2$. 4. Compute derivatives: $$\frac{du}{dx} = 12x^{11}, \quad \frac{dv}{dx} = 2x$$ 5. Apply product rule: $$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x^{12}(2x) + (1 + x^2)(12x^{11})$$ 6. Simplify: $$= 2x^{13} + 12x^{11} + 12x^{13} = 2x^{11}(x^2 + 6(1 + x^2)) = 2x^{11}(7x^2 + 6)$$ --- Exercise 4: 1. Find the derivative of $(x^4 + 7x^2)(5x^5 - 6\sqrt{x})$. - Let $u = x^4 + 7x^2$, $v = 5x^5 - 6x^{1/2}$. - Compute derivatives: $$\frac{du}{dx} = 4x^3 + 14x, \quad \frac{dv}{dx} = 25x^4 - 3x^{-1/2}$$ - Apply product rule: $$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = (x^4 + 7x^2)(25x^4 - 3x^{-1/2}) + (5x^5 - 6x^{1/2})(4x^3 + 14x)$$ 2. Find the derivative of $(x^4\sqrt{x} - 45x^2)(8\sqrt{x})$. - Rewrite: $u = x^{4 + 1/2} - 45x^2 = x^{9/2} - 45x^2$, $v = 8x^{1/2}$. - Compute derivatives: $$\frac{du}{dx} = \frac{9}{2}x^{7/2} - 90x, \quad \frac{dv}{dx} = 4x^{-1/2}$$ - Apply product rule: $$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = (x^{9/2} - 45x^2)(4x^{-1/2}) + (8x^{1/2})(\frac{9}{2}x^{7/2} - 90x)$$ 3. Find the derivative of $(x^{-6} + 57x^2)(5x^{1/3} - 6\sqrt{x})$. - Let $u = x^{-6} + 57x^2$, $v = 5x^{1/3} - 6x^{1/2}$. - Compute derivatives: $$\frac{du}{dx} = -6x^{-7} + 114x, \quad \frac{dv}{dx} = \frac{5}{3}x^{-2/3} - 3x^{-1/2}$$ - Apply product rule: $$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = (x^{-6} + 57x^2)(\frac{5}{3}x^{-2/3} - 3x^{-1/2}) + (5x^{1/3} - 6x^{1/2})(-6x^{-7} + 114x)$$ 4. Find the derivative of $(x^4 + 7x^2)(5x^5 - 6\sqrt{x})(5x + x^2)$. - Let $f = (x^4 + 7x^2)(5x^5 - 6x^{1/2})$, $g = 5x + x^2$. - Use product rule for three factors: $$\frac{d}{dx}(fg) = f' g + f g'$$ - Compute $f'$ from part 1. - Compute $g' = 5 + 2x$. - Then $$\frac{dy}{dx} = f' g + f g'$$ where $f'$ and $f$ are as above. --- Quotient Rule: If $y = \frac{u}{v}$, then $$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ Example: For $y = \frac{2x^3}{4 - 5x^2}$: - Let $u = 2x^3$, $v = 4 - 5x^2$. - Compute derivatives: $$\frac{du}{dx} = 6x^2, \quad \frac{dv}{dx} = -10x$$ - Apply quotient rule: $$\frac{dy}{dx} = \frac{(4 - 5x^2)(6x^2) - (2x^3)(-10x)}{(4 - 5x^2)^2} = \frac{24x^2 - 30x^4 + 20x^4}{(4 - 5x^2)^2} = \frac{24x^2 - 10x^4}{(4 - 5x^2)^2}$$