Step 1: Differentiation of $x$ with respect to $t$

Given $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}$. We apply the Quotient Rule, $\frac{d}{dt}\left[\frac{u}{v}\right] = \frac{v \cdot u' - u \cdot v'}{v^2}$ [Per the Quotient Rule of Calculus].

Let $u = \sin^3 t \implies \frac{du}{dt} = 3\sin^2 t \cos t$ [By Chain Rule].

Let $v = (\cos 2t)^{1/2} \implies \frac{dv}{dt} = \frac{1}{2}(\cos 2t)^{-1/2} \cdot (-\sin 2t) \cdot 2 = \frac{-\sin 2t}{\sqrt{\cos 2t}}$.

$\frac{dx}{dt} = \frac{\sqrt{\cos 2t} \cdot (3\sin^2 t \cos t) - \sin^3 t \cdot \left( \frac{-\sin 2t}{\sqrt{\cos 2t}} \right)}{\cos 2t}$

$\frac{dx}{dt} = \frac{3\sin^2 t \cos t \cos 2t + \sin^3 t \sin 2t}{(\cos 2t)^{3/2}}$

Using $\sin 2t = 2 \sin t \cos t$:

$\frac{dx}{dt} = \frac{\sin^2 t [3 \cos t \cos 2t + 2 \sin^2 t \cos t]}{(\cos 2t)^{3/2}} = \frac{\sin^2 t \cos t [3 \cos 2t + 2 \sin^2 t]}{(\cos 2t)^{3/2}}$

Step 2: Differentiation of $y$ with respect to $t$

Given $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$. Applying Quotient Rule:

Let $u = \cos^3 t \implies \frac{du}{dt} = 3\cos^2 t (-\sin t) = -3\cos^2 t \sin t$.

$\frac{dy}{dt} = \frac{\sqrt{\cos 2t} \cdot (-3\cos^2 t \sin t) - \cos^3 t \cdot \left( \frac{-\sin 2t}{\sqrt{\cos 2t}} \right)}{\cos 2t}$

$\frac{dy}{dt} = \frac{-3\cos^2 t \sin t \cos 2t + \cos^3 t \sin 2t}{(\cos 2t)^{3/2}}$

Using $\sin 2t = 2 \sin t \cos t$:

$\frac{dy}{dt} = \frac{-\cos^2 t \sin t [3 \cos 2t - 2 \cos^2 t]}{(\cos 2t)^{3/2}}$

Step 3: Evaluating the Ratio $\frac{dy}{dx}$

Using the Parametric Differentiation Theorem: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ [Per the Chain Rule for Parametric Equations].

$\frac{dy}{dx} = \frac{- \cos^2 t \sin t [3 \cos 2t - 2 \cos^2 t]}{\sin^2 t \cos t [3 \cos 2t + 2 \sin^2 t]}$

Cancel common terms $\sin t$ and $\cos t$:

$\frac{dy}{dx} = -\frac{\cos t [3(2\cos^2 t - 1) - 2\cos^2 t]}{\sin t [3(1 - 2\sin^2 t) + 2\sin^2 t]}$

$\frac{dy}{dx} = -\frac{\cos t [6\cos^2 t - 3 - 2\cos^2 t]}{\sin t [3 - 6\sin^2 t + 2\sin^2 t]} = -\frac{\cos t [4\cos^2 t - 3]}{\sin t [3 - 4\sin^2 t]}$

Step 4: Trigonometric Simplification

Recall the Triple Angle Identities: $\cos 3t = 4\cos^3 t - 3\cos t$ and $\sin 3t = 3\sin t - 4\sin^3 t$.

$\frac{dy}{dx} = -\frac{\cos t (4\cos^2 t - 3)}{\sin t (3 - 4\sin^2 t)} = -\frac{4\cos^3 t - 3\cos t}{3\sin t - 4\sin^3 t}$

$\frac{dy}{dx} = -\frac{\cos 3t}{\sin 3t} = -\cot 3t$

Final Solution: \frac{dy}{dx} = -\cot 3t

Step 1: Analyzing the Parametric Equations

Given the parametric equations for $x$ and $y$ as functions of parameter $t$:

$x = \sqrt{a^{\sin^{-1} t}} = (a^{\sin^{-1} t})^{1/2} = a^{\frac{1}{2}\sin^{-1} t}$

$y = \sqrt{a^{\cos^{-1} t}} = (a^{\cos^{-1} t})^{1/2} = a^{\frac{1}{2}\cos^{-1} t}$

Step 2: Logarithmic Differentiation

To differentiate exponential functions of the form $a^{f(t)}$, we apply the logarithmic derivative property: $\frac{d}{dt}(a^u) = a^u \ln(a) \cdot \frac{du}{dt}$ [Per the Chain Rule and Derivative of Exponential Functions].

Differentiating $x$ with respect to $t$:

$\frac{dx}{dt} = a^{\frac{1}{2}\sin^{-1} t} \cdot \ln(a) \cdot \frac{d}{dt}\left(\frac{1}{2}\sin^{-1} t\right)$

$\frac{dx}{dt} = a^{\frac{1}{2}\sin^{-1} t} \cdot \ln(a) \cdot \frac{1}{2\sqrt{1-t^2}}$

Differentiating $y$ with respect to $t$:

$\frac{dy}{dt} = a^{\frac{1}{2}\cos^{-1} t} \cdot \ln(a) \cdot \frac{d}{dt}\left(\frac{1}{2}\cos^{-1} t\right)$

$\frac{dy}{dt} = a^{\frac{1}{2}\cos^{-1} t} \cdot \ln(a) \cdot \left(-\frac{1}{2\sqrt{1-t^2}}\right)$

Step 3: Applying the Chain Rule for Parametric Differentiation

By the Chain Rule, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ [Per the theorem on parametric differentiation].

$\frac{dy}{dx} = \frac{a^{\frac{1}{2}\cos^{-1} t} \cdot \ln(a) \cdot \left(-\frac{1}{2\sqrt{1-t^2}}\right)}{a^{\frac{1}{2}\sin^{-1} t} \cdot \ln(a) \cdot \left(\frac{1}{2\sqrt{1-t^2}}\right)}$

Step 4: Algebraic Simplification

Cancel the common terms $\ln(a)$, $\frac{1}{2}$, and $\frac{1}{\sqrt{1-t^2}}$:

$\frac{dy}{dx} = -\frac{a^{\frac{1}{2}\cos^{-1} t}}{a^{\frac{1}{2}\sin^{-1} t}}$

Substitute back the original expressions for $x$ and $y$:

Since $y = a^{\frac{1}{2}\cos^{-1} t}$ and $x = a^{\frac{1}{2}\sin^{-1} t}$:

$\frac{dy}{dx} = -\frac{y}{x}$

Visualization of the Relationship

Final Solution: $\frac{dy}{dx} = -\frac{y}{x}$

Step 1: Differentiation of $x$ with respect to $t$

Given the parametric equation for $x$: $x = a\left(\cos t + \log \tan\left(\frac{t}{2}\right)\right)$

To find $\frac{dx}{dt}$, apply the linearity of the derivative [Per the Sum Rule and Chain Rule]:

$\frac{dx}{dt} = a \left[ \frac{d}{dt}(\cos t) + \frac{d}{dt}\left(\log \tan\frac{t}{2}\right) \right]$

Calculating the individual components:

• $\frac{d}{dt}(\cos t) = -\sin t$
• $\frac{d}{dt}(\log \tan\frac{t}{2}) = \frac{1}{\tan(t/2)} \cdot \sec^2\left(\frac{t}{2}\right) \cdot \frac{1}{2}$ [By the Chain Rule]

Simplifying the second component using trigonometric identities:

$\frac{1}{\tan(t/2)} \cdot \sec^2\left(\frac{t}{2}\right) \cdot \frac{1}{2} = \frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{\cos^2(t/2)} \cdot \frac{1}{2} = \frac{1}{2 \sin(t/2) \cos(t/2)}$

Using the double angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$, we get:

$\frac{1}{\sin t} = \csc t$

Therefore:

$\frac{dx}{dt} = a(-\sin t + \csc t) = a\left(\frac{1 - \sin^2 t}{\sin t}\right) = a\frac{\cos^2 t}{\sin t}$

Step 2: Differentiation of $y$ with respect to $t$

Given the parametric equation for $y$: $y = a \sin t$

Differentiating with respect to $t$:

$\frac{dy}{dt} = a \frac{d}{dt}(\sin t) = a \cos t$ [By the derivative of trigonometric functions]

Step 3: Applying the Parametric Differentiation Rule

To find $\frac{dy}{dx}$, we use the formula [Per the Parametric Chain Rule]:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Substituting the derived expressions:

$\frac{dy}{dx} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}}$

Step 4: Algebraic Simplification

Cancel common factors $a$ and $\cos t$ (assuming $\cos t \neq 0$):

$\frac{dy}{dx} = \frac{\cos t}{\frac{\cos^2 t}{\sin t}} = \cos t \cdot \frac{\sin t}{\cos^2 t} = \frac{\sin t}{\cos t}$

Utilizing the quotient identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$\frac{dy}{dx} = \tan t$

Final Solution: $\frac{dy}{dx} = \tan t$

Given Variables & Initial Setup

We are provided with the parametric equations for $x$ and $y$ as functions of the parameter $\theta$:

• $x = a(\cos \theta + \theta \sin \theta)$
• $y = a(\sin \theta - \theta \cos \theta)$

To find $\frac{dy}{dx}$, we utilize the chain rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ [Per the Rule of Parametric Differentiation].

Step 1: Differentiating $x$ with respect to $\theta$

Given $x = a(\cos \theta + \theta \sin \theta)$, we differentiate with respect to $\theta$ using the Sum Rule and Product Rule:

$\frac{dx}{d\theta} = \frac{d}{d\theta} [a(\cos \theta + \theta \sin \theta)]$

$\frac{dx}{d\theta} = a \left[ \frac{d}{d\theta}(\cos \theta) + \frac{d}{d\theta}(\theta \sin \theta) \right]$

Applying the product rule $\frac{d}{d\theta}(uv) = u\frac{dv}{d\theta} + v\frac{du}{d\theta}$ to the term $\theta \sin \theta$:

$\frac{dx}{d\theta} = a [-\sin \theta + (\theta \cos \theta + \sin \theta \cdot 1)]$

$\frac{dx}{d\theta} = a [-\sin \theta + \theta \cos \theta + \sin \theta]$

$\frac{dx}{d\theta} = a \theta \cos \theta$

Step 2: Differentiating $y$ with respect to $\theta$

Given $y = a(\sin \theta - \theta \cos \theta)$, we differentiate with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta} [a(\sin \theta - \theta \cos \theta)]$

$\frac{dy}{d\theta} = a \left[ \frac{d}{d\theta}(\sin \theta) - \frac{d}{d\theta}(\theta \cos \theta) \right]$

Applying the product rule to the term $\theta \cos \theta$:

$\frac{dy}{d\theta} = a [\cos \theta - (\theta (-\sin \theta) + \cos \theta \cdot 1)]$

$\frac{dy}{d\theta} = a [\cos \theta - (-\theta \sin \theta + \cos \theta)]$

$\frac{dy}{d\theta} = a [\cos \theta + \theta \sin \theta - \cos \theta]$

$\frac{dy}{d\theta} = a \theta \sin \theta$

Step 3: Calculating $\frac{dy}{dx}$

Substitute the results from Step 1 and Step 2 into the parametric derivative formula:

$\frac{dy}{dx} = \frac{a \theta \sin \theta}{a \theta \cos \theta}$

By canceling the common terms $a$ and $\theta$ (where $\theta \neq 0$):

$\frac{dy}{dx} = \frac{\sin \theta}{\cos \theta}$

$\frac{dy}{dx} = \tan \theta$

Final Solution: $\frac{dy}{dx} = \tan \theta$

Step 1: Identifying the Parametric System

We are given two equations where $x$ and $y$ are functions of a common parameter $t$:

• $x = 2at^2$
• $y = at^4$

[Per the Chain Rule for parametric differentiation, if $x = f(t)$ and $y = g(t)$, then $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, provided $\frac{dx}{dt} \neq 0$.]

Step 2: Differentiating $x$ with respect to $t$

Applying the Power Rule of differentiation ($\frac{d}{dt}[t^n] = nt^{n-1}$):
$\frac{dx}{dt} = \frac{d}{dt}(2at^2)$
$\frac{dx}{dt} = 2a \cdot \frac{d}{dt}(t^2)$
$\frac{dx}{dt} = 2a \cdot (2t)$
$\frac{dx}{dt} = 4at$

Step 3: Differentiating $y$ with respect to $t$

Applying the Power Rule of differentiation:
$\frac{dy}{dt} = \frac{d}{dt}(at^4)$
$\frac{dy}{dt} = a \cdot \frac{d}{dt}(t^4)$
$\frac{dy}{dt} = a \cdot (4t^3)$
$\frac{dy}{dt} = 4at^3$

Step 4: Calculating $\frac{dy}{dx}$

Using the relation $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{4at^3}{4at}$
Canceling the common terms $4a$ (assuming $a \neq 0$):
$\frac{dy}{dx} = \frac{t^3}{t}$
Applying the Quotient Rule for exponents ($\frac{t^m}{t^n} = t^{m-n}$):
$\frac{dy}{dx} = t^{3-1}$
$\frac{dy}{dx} = t^2$

Final Solution: $\frac{dy}{dx} = t^2$

Given Variables & Initial Setup

The system is defined by parametric equations where $\theta$ is the parameter:

• $x = a \sec \theta$
• $y = b \tan \theta$

Our objective is to determine $\frac{dy}{dx}$ using the chain rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ [Per the Chain Rule of Calculus].

Step 1: Differentiating $x$ with respect to $\theta$

Given $x = a \sec \theta$. Applying the derivative operator $\frac{d}{d\theta}$ on both sides:

$\frac{dx}{d\theta} = \frac{d}{d\theta}(a \sec \theta)$

Since $a$ is a constant, we extract it and apply the standard derivative of the secant function [Standard Derivative Formula: $\frac{d}{d\theta} \sec \theta = \sec \theta \tan \theta$]:

$\frac{dx}{d\theta} = a \sec \theta \tan \theta$

Step 2: Differentiating $y$ with respect to $\theta$

Given $y = b \tan \theta$. Applying the derivative operator $\frac{d}{d\theta}$ on both sides:

$\frac{dy}{d\theta} = \frac{d}{d\theta}(b \tan \theta)$

Applying the standard derivative of the tangent function [Standard Derivative Formula: $\frac{d}{d\theta} \tan \theta = \sec^2 \theta$]:

$\frac{dy}{d\theta} = b \sec^2 \theta$

Step 3: Calculating $\frac{dy}{dx}$ using the Parametric Quotient

Substitute the results from Step 1 and Step 2 into the quotient formula:

$\frac{dy}{dx} = \frac{b \sec^2 \theta}{a \sec \theta \tan \theta}$

Simplify the expression by canceling the common term $\sec \theta$ [By Algebraic Cancellation Rule]:

$\frac{dy}{dx} = \frac{b \sec \theta}{a \tan \theta}$

Step 4: Algebraic Simplification to Trigonometric Fundamentals

Convert $\sec \theta$ and $\tan \theta$ into sine and cosine functions to simplify further [Trigonometric Identities: $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$]:

$\frac{dy}{dx} = \frac{b}{a} \cdot \frac{1/\cos \theta}{\sin \theta / \cos \theta}$

$\frac{dy}{dx} = \frac{b}{a} \cdot \frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta}$

$\frac{dy}{dx} = \frac{b}{a \sin \theta}$

Applying the reciprocal identity $\frac{1}{\sin \theta} = \csc \theta$:

$\frac{dy}{dx} = \frac{b}{a} \csc \theta$

Final Solution: $\frac{dy}{dx} = \frac{b}{a} \csc \theta$

Step 1: Identifying the Parametric System

We are given the parametric equations where $\theta$ acts as the independent parameter:

$x = a \cos \theta$

$y = b \cos \theta$

To find the derivative $\frac{dy}{dx}$, we employ the Chain Rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ [Per the Chain Rule for inverse functions, provided $\frac{dx}{d\theta} \neq 0$].

Step 2: Differentiating $x$ with respect to $\theta$

Given $x = a \cos \theta$, differentiate both sides with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta} (a \cos \theta)$

Using the constant multiple rule and the standard derivative $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$:

$\frac{dx}{d\theta} = a (-\sin \theta) = -a \sin \theta$

Step 3: Differentiating $y$ with respect to $\theta$

Given $y = b \cos \theta$, differentiate both sides with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta} (b \cos \theta)$

Applying the same derivative rule:

$\frac{dy}{d\theta} = b (-\sin \theta) = -b \sin \theta$

Step 4: Constructing the Derivative $\frac{dy}{dx}$

Substitute the results from Step 2 and Step 3 into the quotient $\frac{dy/d\theta}{dx/d\theta}$:

$\frac{dy}{dx} = \frac{-b \sin \theta}{-a \sin \theta}$

Step 5: Simplification

Assuming $\sin \theta \neq 0$, the trigonometric terms cancel out:

$\frac{dy}{dx} = \frac{-b}{-a} \cdot \frac{\sin \theta}{\sin \theta}$

$\frac{dy}{dx} = \frac{b}{a}$

Final Solution: $\frac{dy}{dx} = \frac{b}{a}$

Step 1: Identifying the Parametric Equations

Given the parametric equations for the variables $x$ and $y$ with respect to the parameter $t$:

• $x = \sin t$
• $y = \cos 2t$

Step 2: Differentiating $x$ with respect to $t$

To find the derivative of $x$ with respect to $t$, we apply the standard derivative rule for trigonometric functions [Per the derivative formula $\frac{d}{dt}(\sin t) = \cos t$]:

$\frac{dx}{dt} = \frac{d}{dt}(\sin t) = \cos t$

Step 3: Differentiating $y$ with respect to $t$

To find the derivative of $y$ with respect to $t$, we apply the Chain Rule [Per the Chain Rule: $\frac{d}{dt}f(g(t)) = f'(g(t)) \cdot g'(t)$]:

$\frac{dy}{dt} = \frac{d}{dt}(\cos 2t)$

Since $\frac{d}{du}(\cos u) = -\sin u$ and $\frac{d}{dt}(2t) = 2$:

$\frac{dy}{dt} = -\sin(2t) \cdot \frac{d}{dt}(2t) = -2\sin 2t$

Step 4: Applying the Parametric Derivative Formula

To obtain $\frac{dy}{dx}$, we use the formula for parametric differentiation [Per the Quotient Rule for derivatives]:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Substituting the derivatives found in Steps 2 and 3:

$\frac{dy}{dx} = \frac{-2\sin 2t}{\cos t}$

Step 5: Simplification using Trigonometric Identities

We utilize the double-angle identity for sine [Per the identity: $\sin 2\theta = 2\sin \theta \cos \theta$]:

$\frac{dy}{dx} = \frac{-2(2\sin t \cos t)}{\cos t}$

Assuming $\cos t \neq 0$, we cancel the term $\cos t$ from the numerator and denominator:

$\frac{dy}{dx} = -4\sin t$

Visual Representation of the Parametric Slope

Final Solution: \frac{dy}{dx} = -4\sin t

Step 1: Given Parametric Equations

The system is defined by the parameter $t$ where $t \in \mathbb{R} \setminus \{0\}$. The equations are:

• $x = 4t$
• $y = 4t^{-1}$

Step 2: Differentiating $x$ with respect to $t$

To find the derivative of the independent variable $x$ with respect to the parameter $t$:

$\frac{dx}{dt} = \frac{d}{dt}(4t)$

Applying the Power Rule of differentiation, $\frac{d}{dt}(at^n) = ant^{n-1}$ [Per the Power Rule of Calculus]:

$\frac{dx}{dt} = 4(1) = 4$

Step 3: Differentiating $y$ with respect to $t$

To find the derivative of the dependent variable $y$ with respect to the parameter $t$:

$\frac{dy}{dt} = \frac{d}{dt}(4t^{-1})$

Applying the Power Rule, $\frac{d}{dt}(ct^n) = cnt^{n-1}$:

$\frac{dy}{dt} = 4(-1)t^{-1-1}$

$\frac{dy}{dt} = -4t^{-2} = -\frac{4}{t^2}$

Step 4: Applying the Chain Rule for Parametric Differentiation

To determine $\frac{dy}{dx}$ without eliminating the parameter, we utilize the relationship between parametric derivatives [Per the Chain Rule for Parametric Equations]:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Substituting the derived expressions from Step 2 and Step 3:

$\frac{dy}{dx} = \frac{-\frac{4}{t^2}}{4}$

Step 5: Algebraic Simplification

Simplifying the complex fraction:

$\frac{dy}{dx} = -\frac{4}{t^2} \cdot \frac{1}{4}$

The constant $4$ cancels out:

$\frac{dy}{dx} = -\frac{1}{t^2}$

Visual Representation of the Curve

Final Solution: \frac{dy}{dx} = -\frac{1}{t^2}

Step 1: Identifying the Parametric Equations

The variables $x$ and $y$ are functions of the parameter $\theta$:

• $x = \cos \theta - \cos 2\theta$
• $y = \sin \theta - \sin 2\theta$

To find $\frac{dy}{dx}$, we apply the Chain Rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ [Per the Chain Rule for Parametric Differentiation].

Step 2: Differentiating $x$ with respect to $\theta$

Differentiating $x = \cos \theta - \cos 2\theta$ with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}(\cos \theta) - \frac{d}{d\theta}(\cos 2\theta)$

Applying the standard derivative $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$ and the Chain Rule $\frac{d}{d\theta}(\cos 2\theta) = -\sin 2\theta \cdot \frac{d}{d\theta}(2\theta) = -2\sin 2\theta$ [By Rule of Chain Rule]:

$\frac{dx}{d\theta} = -\sin \theta - (-2 \sin 2\theta)$

$\frac{dx}{d\theta} = 2 \sin 2\theta - \sin \theta$

Step 3: Differentiating $y$ with respect to $\theta$

Differentiating $y = \sin \theta - \sin 2\theta$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta}(\sin \theta) - \frac{d}{d\theta}(\sin 2\theta)$

Applying the standard derivative $\frac{d}{d\theta}(\sin \theta) = \cos \theta$ and the Chain Rule $\frac{d}{d\theta}(\sin 2\theta) = \cos 2\theta \cdot 2$:

$\frac{dy}{d\theta} = \cos \theta - 2 \cos 2\theta$

Step 4: Calculating $\frac{dy}{dx}$

Substitute the results from Step 2 and Step 3 into the quotient expression:

$\frac{dy}{dx} = \frac{\cos \theta - 2 \cos 2\theta}{2 \sin 2\theta - \sin \theta}$

Visual Representation of Parametric Slope

Step 5: Final Simplification (Optional)

While the expression is complete, it is mathematically standard to present the result in its derived form as further trigonometric expansion using double-angle identities ($\cos 2\theta = 2\cos^2\theta - 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$) does not lead to significant further reduction in this specific rational function.

Final Solution: $\frac{dy}{dx} = \frac{\cos \theta - 2 \cos 2\theta}{2 \sin 2\theta - \sin \theta}$

Step 1: Analyzing the Parametric Equations

The system is defined by the parameter $\theta$ as follows:

• $x = a(\theta - \sin \theta)$
• $y = a(1 + \cos \theta)$

[Per the Chain Rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$, provided $dx/d\theta \neq 0$.]

Step 2: Differentiating $x$ with respect to $\theta$

Applying the linearity of the derivative and the standard derivative formula $\frac{d}{d\theta}(\sin \theta) = \cos \theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta} [a(\theta - \sin \theta)]$

$\frac{dx}{d\theta} = a \left( \frac{d}{d\theta}(\theta) - \frac{d}{d\theta}(\sin \theta) \right)$

$\frac{dx}{d\theta} = a(1 - \cos \theta)$

Step 3: Differentiating $y$ with respect to $\theta$

Applying the standard derivative formula $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta} [a(1 + \cos \theta)]$

$\frac{dy}{d\theta} = a \left( \frac{d}{d\theta}(1) + \frac{d}{d\theta}(\cos \theta) \right)$

$\frac{dy}{d\theta} = a(0 - \sin \theta) = -a \sin \theta$

Step 4: Calculating $\frac{dy}{dx}$

Substitute the results from Step 2 and Step 3 into the parametric derivative formula:

$\frac{dy}{dx} = \frac{-a \sin \theta}{a(1 - \cos \theta)}$

$\frac{dy}{dx} = \frac{-\sin \theta}{1 - \cos \theta}$

Step 5: Trigonometric Simplification

Utilize the half-angle identities to simplify the expression [Identity: $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 - \cos \theta = 2 \sin^2(\theta/2)$]:

$\frac{dy}{dx} = \frac{-2 \sin(\theta/2) \cos(\theta/2)}{2 \sin^2(\theta/2)}$

Cancel the common terms $2$ and $\sin(\theta/2)$:

$\frac{dy}{dx} = -\frac{\cos(\theta/2)}{\sin(\theta/2)}$

Since $\cot \theta = \frac{\cos \theta}{\sin \theta}$:

$\frac{dy}{dx} = -\cot\left(\frac{\theta}{2}\right)$

Final Solution: $\frac{dy}{dx} = -\cot\left(\frac{\theta}{2}\right)$