已知 $y = e^{\cos^{-1} x}$。

1. Find the first derivative of $y$ with respect to $x$ using the chain rule.

$y = e^{\cos^{-1} x}$

Differentiating with respect to $x$:
$\frac{dy}{dx} = e^{\cos^{-1} x} \cdot \frac{d}{dx}(\cos^{-1} x)$
$\frac{dy}{dx} = e^{\cos^{-1} x} \left( -\frac{1}{\sqrt{1 - x^2}} \right)$

Since $y = e^{\cos^{-1} x}$, this can be written as:
$\frac{dy}{dx} = -\frac{y}{\sqrt{1 - x^2}}$

2. Rearrange the equation to prepare for implicit differentiation without using the quotient rule.

Multiply both sides by $\sqrt{1 - x^2}$:
$\sqrt{1 - x^2} \frac{dy}{dx} = -y$

3. Differentiate implicitly with respect to $x$ using the product rule.

Applying the product rule on the left hand side:
$\sqrt{1 - x^2} \frac{d^2y}{dx^2} + \frac{dy}{dx} \frac{d}{dx}\left(\sqrt{1 - x^2}\right) = -\frac{dy}{dx}$
$\sqrt{1 - x^2} \frac{d^2y}{dx^2} + \frac{dy}{dx} \left( \frac{1}{2\sqrt{1 - x^2}} (-2x) \right) = -\frac{dy}{dx}$
$\sqrt{1 - x^2} \frac{d^2y}{dx^2} - \frac{x}{\sqrt{1 - x^2}} \frac{dy}{dx} = -\frac{dy}{dx}$

4. Simplify the equation to reach the target format.

Multiply the entire equation by $\sqrt{1 - x^2}$ to clear the denominator:
$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -\sqrt{1 - x^2} \frac{dy}{dx}$

From Step 2, substitute $-y = \sqrt{1 - x^2} \frac{dy}{dx}$ (which means $-\sqrt{1 - x^2} \frac{dy}{dx} = y$):
$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = y \quad \text{(shown)}$

1. Evaluate the function and its first derivative at $x = 0$.

$y = e^{\cos^{-1} x}$
At $x = 0$, $y = e^{\cos^{-1} 0} = e^{\frac{\pi}{2}}$.

$\dfrac{dy}{dx} = e^{\cos^{-1} x} \cdot \left(-\dfrac{1}{\sqrt{1 - x^2}}\right)$
At $x = 0$, $\dfrac{dy}{dx} = e^{\frac{\pi}{2}} \cdot \left(-\dfrac{1}{1}\right) = -e^{\frac{\pi}{2}}$.

2. Use the differential equation from part (i) to find the second derivative at $x = 0$.

From (i), $(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} = y$

Substituting $x = 0$:
$(1 - 0)\dfrac{d^2 y}{dx^2} - 0 = e^{\frac{\pi}{2}}$
$\dfrac{d^2 y}{dx^2} = e^{\frac{\pi}{2}}$

3. Differentiate the differential equation implicitly to find the third derivative at $x = 0$.

Differentiating $(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} = y$ with respect to $x$ using the product rule:
$\left[ (1 - x^2)\dfrac{d^3 y}{dx^3} + (-2x)\dfrac{d^2 y}{dx^2} \right] - \left[ x\dfrac{d^2 y}{dx^2} + (1)\dfrac{dy}{dx} \right] = \dfrac{dy}{dx}$

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

Substituting $x = 0$:
$(1)\dfrac{d^3 y}{dx^3} - 0 - 2\left(-e^{\frac{\pi}{2}}\right) = 0$
$\dfrac{d^3 y}{dx^3} = -2e^{\frac{\pi}{2}}$

4. Form the Maclaurin's series up to the term in $x^3$.

$y = y(0) + y'(0)x + \dfrac{y''(0)}{2!}x^2 + \dfrac{y'''(0)}{3!}x^3 + \dots$
$y = e^{\frac{\pi}{2}} + \left(-e^{\frac{\pi}{2}}\right)x + \dfrac{e^{\frac{\pi}{2}}}{2}x^2 + \dfrac{-2e^{\frac{\pi}{2}}}{6}x^3 + \dots$
$y = e^{\frac{\pi}{2}} - e^{\frac{\pi}{2}}x + \dfrac{e^{\frac{\pi}{2}}}{2}x^2 - \dfrac{e^{\frac{\pi}{2}}}{3}x^3 + \dots$

1. Determine the Maclaurin series for $y$ up to the term in $x^3$ as required from part (ii).

Given $y = e^{\cos^{-1} x}$. At $x = 0$, $y = e^{\pi/2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = -\frac{e^{\cos^{-1} x}}{\sqrt{1-x^2}} \implies y'(0) = -e^{\pi/2}$
From part (i), $(1 - x^2)y'' - xy' = y$. At $x = 0$, $y''(0) = y(0) = e^{\pi/2}$.
Differentiating the equation from (i) with respect to $x$:
$-2xy'' + (1 - x^2)y''' - y' - xy'' = y'$
$(1 - x^2)y''' - 3xy'' - 2y' = 0$
At $x = 0$, $y'''(0) - 2(-e^{\pi/2}) = 0 \implies y'''(0) = -2e^{\pi/2}$.
Thus, the Maclaurin series is $y \approx e^{\pi/2} - e^{\pi/2}x + \frac{e^{\pi/2}}{2}x^2 - \frac{2e^{\pi/2}}{6}x^3 = e^{\pi/2}\left(1 - x + \frac{x^2}{2} - \frac{x^3}{3}\right)$.

2. Find the expansion of $-\frac{e^{\cos^{-1} x}}{\sqrt{1 - x^2}}$ by differentiating the Maclaurin series for $y$.

Observe that the required expression is exactly $\frac{dy}{dx}$.
Using the expansion of $y$ from part (ii):
$\frac{dy}{dx} \approx \frac{d}{dx} \left[ e^{\pi/2}\left(1 - x + \frac{x^2}{2} - \frac{x^3}{3}\right) \right]$
$\frac{dy}{dx} \approx e^{\pi/2}\left(-1 + x - x^2\right)$

3. Estimate the gradient of the tangent at $x = 0.5$.

The gradient of the tangent at $x = 0.5$ is given by the value of $\frac{dy}{dx}$ at $x = 0.5$.
Substituting $x = 0.5$ into the expansion for $\frac{dy}{dx}$:
$\text{Gradient } \approx e^{\pi/2}\left(-1 + 0.5 - (0.5)^2\right)$
$= e^{\pi/2}(-1 + 0.5 - 0.25)$
$= -0.75e^{\pi/2} = -\frac{3}{4}e^{\pi/2}$