Given the equation of the curve of the curve defined by y=x³ − 6x² + 9x − 2, Determine: (i) The critical points, (ii) Their nature, (iii) Hence sketch the curve.

$$\frac{dy}{du} = \frac{d}{du}(u^4) = 4u^3$$ $$\frac{du}{dx} = \frac{d}{dx}(3x^2-5x+8) = 6x-5$$

$$\frac{dy}{dx} = 4u^3 (6x-5)$$

Substitute $u = 3x^2-5x+8$ back into the expression.

$$\boxed{\frac{dy}{dx} = 4(3x^2-5x+8)^3 (6x-5)}$$

a) (ii) To find the derivative of $y = e^{\cos x}$ using the chain rule:
Let $u = \cos x$. Then $y = e^u$.

$$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u$$ $$\frac{du}{dx} = \frac{d}{dx}(\cos x) = -\sin x$$

$$\frac{dy}{dx} = e^u (-\sin x)$$

Substitute $u = \cos x$ back into the expression.

$$\boxed{\frac{dy}{dx} = -e^{\cos x} \sin x}$$

a) (iii) To find the derivative of $y = \ln(x^4+3)$ using the chain rule:
Let $u = x^4+3$. Then $y = \ln u$.

$$\frac{dy}{du} = \frac{d}{du}(\ln u) = \frac{1}{u}$$ $$\frac{du}{dx} = \frac{d}{dx}(x^4+3) = 4x^3$$

$$\frac{dy}{dx} = \frac{1}{u} (4x^3)$$

Substitute $u = x^4+3$ back into the expression.

$$\boxed{\frac{dy}{dx} = \frac{4x^3}{x^4+3}}$$

b) Given $y = e^{3x}$, show that $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 12y = 0$.

$$y = e^{3x}$$ $$\frac{dy}{dx} = \frac{d}{dx}(e^{3x}) = 3e^{3x}$$

$$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 12y = 0$$ $$(9e^{3x}) + (3e^{3x}) - 12(e^{3x}) = 0$$ $$9e^{3x} + 3e^{3x} - 12e^{3x} = 0$$ $$12e^{3x} - 12e^{3x} = 0$$ $$0 = 0$$

Since the left side equals the right side, the statement is proven.