Given that $y = \cos\{\ln(1 + x)\}$, prove that

1. Apply the chain rule to differentiate $y$ with respect to $x$.

$\frac{dy}{dx} = -\sin\{\ln(1 + x)\} \cdot \frac{d}{dx}[\ln(1 + x)]$
$\frac{dy}{dx} = -\sin\{\ln(1 + x)\} \cdot \frac{1}{1 + x}$

2. Rearrange the equation to obtain the required result.

$(1 + x)\frac{dy}{dx} = -\sin\{\ln(1 + x)\}$

1. Differentiate the equation from part (i) implicitly with respect to $x$, using the product and chain rules.

$\frac{d}{dx} \left[ (1 + x)\frac{dy}{dx} \right] = \frac{d}{dx} \left[ -\sin\{\ln(1 + x)\} \right]$
$(1)\frac{dy}{dx} + (1 + x)\frac{d^2y}{dx^2} = -\cos\{\ln(1 + x)\} \cdot \frac{1}{1 + x}$

2. Multiply by $(1+x)$, substitute $y = \cos\{\ln(1+x)\}$, and rearrange to match the target equation.

Multiply both sides by $(1 + x)$:
$(1 + x)\frac{dy}{dx} + (1 + x)^2\frac{d^2y}{dx^2} = -\cos\{\ln(1 + x)\}$
Substitute the original function $y = \cos\{\ln(1 + x)\}$ into the right-hand side:
$(1 + x)\frac{dy}{dx} + (1 + x)^2\frac{d^2y}{dx^2} = -y$
Rearrange to obtain the target equation:
$(1 + x)^2\frac{d^2y}{dx^2} + (1 + x)\frac{dy}{dx} + y = 0 \quad \text{(shown)}$

1. Differentiate the equation from part (ii) with respect to $x$ using the product rule.

$\dfrac{d}{dx}\left[ (1 + x)^2\dfrac{d^2y}{dx^2} + (1 + x)\dfrac{dy}{dx} + y \right] = 0$

Applying the product rule to the first two terms:
$\left[ (1 + x)^2\dfrac{d^3y}{dx^3} + 2(1 + x)\dfrac{d^2y}{dx^2} \right] + \left[ (1 + x)\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} \right] + \dfrac{dy}{dx} = 0$

2. Group like terms to simplify the expression into the final equation.

$(1+x)^2\dfrac{d^3y}{dx^3} + \left[2(1+x) + (1+x)\right]\dfrac{d^2y}{dx^2} + \left[1 + 1\right]\dfrac{dy}{dx} = 0

(1+x)^2\dfrac{d^3y}{dx^3} + 3(1+x)\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} = 0$

1. Derive the equation relating the first, second, and third derivatives as required from part (iii).

Differentiating the result from (ii), $(1 + x)^2\dfrac{d^2y}{dx^2} + (1 + x)\dfrac{dy}{dx} + y = 0$, implicitly with respect to $x$:

$(1 + x)^2\dfrac{d^3y}{dx^3} + 2(1 + x)\dfrac{d^2y}{dx^2} + (1 + x)\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} + \dfrac{dy}{dx} = 0$

$(1 + x)^2\dfrac{d^3y}{dx^3} + 3(1 + x)\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} = 0$

2. Evaluate the function and its derivatives at $x = 0$.

Substitute $x = 0$ into the original function and the respective derivative equations:

$y(0) = \cos(\ln 1) = \cos(0) = 1$

From (i):
$(1 + 0)\dfrac{dy}{dx}\bigg|_{x=0} = -\sin(\ln 1) = 0 \implies \dfrac{dy}{dx}\bigg|_{x=0} = 0$

From (ii):
$(1 + 0)^2\dfrac{d^2y}{dx^2}\bigg|_{x=0} + (1 + 0)(0) + 1 = 0 \implies \dfrac{d^2y}{dx^2}\bigg|_{x=0} = -1$

From (iii):
$(1 + 0)^2\dfrac{d^3y}{dx^3}\bigg|_{x=0} + 3(1 + 0)(-1) + 2(0) = 0 \implies \dfrac{d^3y}{dx^3}\bigg|_{x=0} = 3$

3. Construct the Maclaurin series using the standard formula.

Using the Maclaurin series expansion formula $y = y(0) + y'(0)x + \dfrac{y''(0)}{2!}x^2 + \dfrac{y'''(0)}{3!}x^3 + \ldots$

$y = 1 + (0)x + \dfrac{-1}{2!}x^2 + \dfrac{3}{3!}x^3 + \ldots$

$y = 1 - \dfrac{1}{2}x^2 + \dfrac{1}{2}x^3$

1. State the standard Maclaurin series for $\ln(1+x)$ and $\cos u$.

From the standard series expansions, we have:
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$
$\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots$

2. Substitute $u = \ln(1+x)$ into the expansion for $\cos u$ and expand up to the term in $x^3$.

Let $u = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$.
To find the expansion of $\cos u$ up to $x^3$, we first evaluate $u^2$ up to $x^3$:
$u^2 = \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots\right)^2$
$u^2 = x^2 + 2(x)\left(-\frac{x^2}{2}\right) + \dots$
$u^2 = x^2 - x^3 + \dots$

Substituting this into the cosine expansion gives:
$\cos(\ln(1+x)) = 1 - \frac{1}{2}\left(x^2 - x^3 + \dots\right)$
$y = 1 - \frac{1}{2}x^2 + \frac{1}{2}x^3 + \dots$

3. Conclude the verification.

The series obtained up to the term in $x^3$ is $1 - \frac{1}{2}x^2 + \frac{1}{2}x^3$, which is precisely the same result as the Maclaurin series derived via differentiation in part (iv).