Here are the derivatives of the given functions. For the first function, $y = \frac{1}{2}x^4$, we will differentiate from the first principle. The first principle of differentiation is given by: $$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ Step 1: Identify $f(x)$ and $f(x+h)$. Given $f(x) = \frac{1}{2}x^4$. Then $f(x+h) = \frac{1}{2}(x+h)^4$. Step 2: Expand $(x+h)^4$. $$(x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4$$ Step 3: Substitute the expansion into $f(x+h)$. $$f(x+h) = \frac{1}{2}(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4)$$ $$f(x+h) = \frac{1}{2}x^4 + 2x^3h + 3x^2h^2 + 2xh^3 + \frac{1}{2}h^4$$ Step 4: Calculate $f(x+h) - f(x)$. $$f(x+h) - f(x) = \left(\frac{1}{2}x^4 + 2x^3h + 3x^2h^2 + 2xh^3 + \frac{1}{2}h^4\right) - \frac{1}{2}x^4$$ $$f(x+h) - f(x) = 2x^3h + 3x^2h^2 + 2xh^3 + \frac{1}{2}h^4$$ Step 5: Divide by $h$. $$\frac{f(x+h) - f(x)}{h} = \frac{2x^3h + 3x^2h^2 + 2xh^3 + \frac{1}{2}h^4}{h}$$ $$\frac{f(x+h) - f(x)}{h} = 2x^3 + 3x^2h + 2xh^2 + \frac{1}{2}h^3$$ Step 6: Take the limit as $h \to 0$. $$\frac{dy}{dx} = \lim_{h \to 0} (2x^3 + 3x^2h + 2xh^2 + \frac{1}{2}h^3)$$ $$\frac{dy}{dx} = 2x^3 + 3x^2(0) + 2x(0)^2 + \frac{1}{2}(0)^3$$ $$\frac{dy}{dx} = 2x^3$$ The derivative of the first function is: $$\boxed{\frac{dy}{dx} = 2x^3}$$ For the second function, $y = \sqrt{x}(1+2x)^2$: Step 1: Rewrite the function using exponent notation. $$y = x^{1/2}(1+2x)^2$$ Step 2: Apply the product rule, $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$. Let $u = x^{1/2}$ and $v = (1+2x)^2$. Step 3: Find $\frac{du}{dx}$. $$\frac{du}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$$ Step 4: Find $\frac{dv}{dx}$ using the chain rule. Let $w = 1+2x$, so $v = w^2$. $$\frac{dv}{dw} = 2w$$ $$\frac{dw}{dx} = \frac{d}{dx}(1+2x) = 2$$ $$\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx} = 2w \cdot 2 = 4w = 4(1+2x)$$ Step 5: Substitute $u, v, \frac{du}{dx}, \frac{dv}{dx}$ into the product rule formula. $$\frac{dy}{dx} = x^{1/2} \cdot 4(1+2x) + (1+2x)^2 \cdot \frac{1}{2}x^{-1/2}$$ $$\frac{dy}{dx} = 4\sqrt{x}(1+2x) + \frac{(1+2x)^2}{2\sqrt{x}}$$ Step 6: Simplify the expression by finding a common denominator. $$\frac{dy}{dx} = \frac{4\sqrt{x}(1+2x) \cdot 2\sqrt{x}}{2\sqrt{x}} + \frac{(1+2x)^2}{2\sqrt{x}}$$ $$\frac{dy}{dx} = \frac{8x(1+2x) + (1+2x)^2}{2\sqrt{x}}$$ Step 7: Factor out $(1+2x)$ from the numerator. $$\frac{dy}{dx} = \frac{(1+2x)[8x + (1+2x)]}{2\sqrt{x}}$$ $$\frac{dy}{dx} = \frac{(1+2x)(8x + 1 + 2x)}{2\sqrt{x}}$$ $$\frac{dy}{dx} = \frac{(1+2x)(10x + 1)}{2\sqrt{x}}$$ The derivative of the second function is: $$\boxed{\frac{dy}{dx} = \frac{(1+2x)(10x + 1)}{2\sqrt{x}}}$$ That's 2 down. 3 left today — send the next one.