Differentiate the following functions: (i) y=e^x sin 4x (ii) y = (x²+1)/(x²+x+1). Given the equation of a curve: xy² +3y−4x=7. Find dy/dx.

$$u' = \frac{d}{dx}(e^x) = e^x$$ $$v' = \frac{d}{dx}(\sin(4x)) = \cos(4x) \cdot \frac{d}{dx}(4x) = 4\cos(4x)$$

$$\frac{dy}{dx} = u'v + uv' = (e^x)(\sin(4x)) + (e^x)(4\cos(4x))$$ $$\frac{dy}{dx} = e^x \sin(4x) + 4e^x \cos(4x)$$ $$\boxed{\frac{dy}{dx} = e^x (\sin(4x) + 4\cos(4x))}$$

a) (ii) To differentiate $y = \frac{x^2+1}{x^2+x+1}$, we use the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.
Let $u = x^2+1$ and $v = x^2+x+1$.

$$u' = \frac{d}{dx}(x^2+1) = 2x$$ $$v' = \frac{d}{dx}(x^2+x+1) = 2x+1$$

$$\text{Numerator} = (2x^3+2x^2+2x) - (2x^3+x^2+2x+1)$$ $$\text{Numerator} = 2x^3+2x^2+2x - 2x^3-x^2-2x-1$$ $$\text{Numerator} = x^2-1$$

$$\boxed{\frac{dy}{dx} = \frac{x^2-1}{(x^2+x+1)^2}}$$

b) To find $\frac{dy}{dx}$ for the curve $xy^2 + 3y - 4x = 7$, we use implicit differentiation.

$$\frac{d}{dx}(xy^2) + \frac{d}{dx}(3y) - \frac{d}{dx}(4x) = \frac{d}{dx}(7)$$

Using the product rule for $xy^2$: $\frac{d}{dx}(xy^2) = (1)y^2 + x(2y\frac{dy}{dx}) = y^2 + 2xy\frac{dy}{dx}$.

$$y^2 + 2xy\frac{dy}{dx} + 3\frac{dy}{dx} - 4 = 0$$