Evaluate all the following: i) Lim 2x+3 / 4x+3 ii) Lim x->0 x/(x2+2) iii) y=(3x3−8x2−1)6 iv) 4xy+y2 = x2+2

As $x \to \infty$, $\frac{3}{x} \to 0$ and $\frac{1}{x} \to 0$.

$$\frac{2+0}{8+0} = \frac{2}{8}$$

$$\frac{2}{8} = \frac{1}{4}$$

The limit is $\boxed{\frac{1}{4}}$.

ii) $\lim_{x \to 0} \frac{2x^2+8x+3}{x^2+2}$

$$\frac{0+0+3}{0+2} = \frac{3}{2}$$

The limit is $\boxed{\frac{3}{2}}$.

Question 2: Find $\frac{dy}{dx}$ of the following

i) $y = \frac{3x^2-8}{4x+3}$

Then $u' = 6x$ and $v' = 4$.

$$\frac{dy}{dx} = \frac{(6x)(4x+3) - (3x^2-8)(4)}{(4x+3)^2}$$

$$\frac{dy}{dx} = \frac{24x^2+18x - 12x^2+32}{(4x+3)^2} = \frac{12x^2+18x+32}{(4x+3)^2}$$

The derivative is $\boxed{\frac{12x^2+18x+32}{(4x+3)^2}}$.

ii) $y = (3x^3-8x^2-1)^6$

First, find $f'(x)$: $f'(x) = \frac{d}{dx}(3x^3-8x^2-1) = 9x^2-16x$.

$$\frac{dy}{dx} = 6(3x^3-8x^2-1)^{6-1} \cdot (9x^2-16x)$$

$$\frac{dy}{dx} = 6(9x^2-16x)(3x^3-8x^2-1)^5$$

The derivative is $\boxed{6(9x^2-16x)(3x^3-8x^2-1)^5}$.

iii) $4xy + y^2 = x^2+2$

Remember to use the product rule for $4xy$ and the chain rule for $y^2$.

$$\frac{d}{dx}(4xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(x^2) + \frac{d}{dx}(2)$$ $$(4 \cdot y + 4x \cdot \frac{dy}{dx}) + (2y \cdot \frac{dy}{dx}) = 2x + 0$$ $$4y + 4x\frac{dy}{dx} + 2y\frac{dy}{dx} = 2x$$

$$\frac{dy}{dx} = \frac{2(x - 2y)}{2(2x + y)} = \frac{x - 2y}{2x + y}$$

The derivative is $\boxed{\frac{x - 2y}{2x + y}}$.