The problem consists of three parts: (i) Differentiate $y = (3x+1)^3$ with respect to $x$. (ii) Given $f(x) = \frac{x}{x+2}$ and $g(x) = 2x-1$, find (a) $(f \circ g)(x)$ and (b) $(g \circ f)(\frac{3}{2})$. (iii) Rationalize the expression $\frac{3}{\sqrt{2}-1}$.

1. Problem Description

The problem consists of three parts:

(i) Differentiate $y = (3x+1)^3$ with respect to $x$.

(ii) Given $f(x) = \frac{x}{x+2}$ and $g(x) = 2x-1$, find (a) $(f \circ g)(x)$ and (b) $(g \circ f)(\frac{3}{2})$.

(iii) Rationalize the expression $\frac{3}{\sqrt{2}-1}$.

2. Solution Steps

(i) Differentiate $y = (3x+1)^3$ with respect to $x$.

We use the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. Let $u = 3x+1$. Then $y = u^3$.

$\frac{dy}{du} = 3u^2$

$\frac{du}{dx} = 3$

$\frac{dy}{dx} = 3u^2 \cdot 3 = 9u^2 = 9(3x+1)^2$

$\frac{dy}{dx} = 9(9x^2 + 6x + 1) = 81x^2 + 54x + 9$

(ii)

(a) $(f \circ g)(x) = f(g(x)) = f(2x-1) = \frac{2x-1}{(2x-1)+2} = \frac{2x-1}{2x+1}$

(b) $(g \circ f)(x) = g(f(x)) = g(\frac{x}{x+2}) = 2(\frac{x}{x+2}) - 1 = \frac{2x}{x+2} - 1 = \frac{2x - (x+2)}{x+2} = \frac{x-2}{x+2}$.

$(g \circ f)(\frac{3}{2}) = \frac{\frac{3}{2} - 2}{\frac{3}{2} + 2} = \frac{\frac{3-4}{2}}{\frac{3+4}{2}} = \frac{-1}{7} = -\frac{1}{7}$

(iii) Rationalize $\frac{3}{\sqrt{2}-1}$.

Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2}+1$.

$\frac{3}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{3(\sqrt{2}+1)}{(\sqrt{2})^2 - 1^2} = \frac{3(\sqrt{2}+1)}{2-1} = \frac{3(\sqrt{2}+1)}{1} = 3(\sqrt{2}+1) = 3\sqrt{2} + 3$

3. Final Answer

(i) $\frac{dy}{dx} = 81x^2 + 54x + 9$

(ii) (a) $(f \circ g)(x) = \frac{2x-1}{2x+1}$

(b) $(g \circ f)(\frac{3}{2}) = -\frac{1}{7}$

(iii) $3\sqrt{2}+3$