Q.Find $\frac{dy}{dx}$ in the following: $x = \cos \theta - \cos 2\theta, y = \sin \theta - \sin 2\theta$

We are given two parametric equations in $\theta$, so we use the chain rule:

First, differentiate each with respect to $\theta$:

Therefore, …

When you see $x$ and $y$ both given in terms of a third variable (here $\theta$), you’re looking at a parametric curve. The question asks for $\frac{dy}{dx}$ — the slope of the curve at any point. But you can’t directly differentiate $y$ with respect to $x$ because they aren’t written as $y = f(x)$. Instead, you use the chain rule in a clever way.

The core idea:

If $x = f(\theta)$ and $y = g(\theta)$, then

provided $dx/d\theta \neq 0$. This works because $dy/d\theta = (dy/dx) \cdot (dx/d\theta)$ by the chain rule, so you just divide through.

Let’s apply it step by step.

1. Differentiate $x$ with respect to $\theta$ $x = \cos\theta - \cos 2\theta$ The derivative of $\cos\theta$ is $-\sin\theta$. The derivative of $\cos 2\theta$ is $-\sin 2\theta \cdot 2 = -2\sin 2\theta$ (chain rule). So:

2. Differentiate $y$ with respect to $\theta$ $y = \sin\theta - \sin 2\theta$ Derivative of $\sin\theta$ is $\cos\theta$. Derivative of $\sin 2\theta$ is $\cos 2\theta \cdot 2 = 2\cos 2\theta$. So:

3. Form the ratio …

Method: Differentiating a Parametric Curve

This method applies whenever a curve is given through a third variable (a parameter — commonly $t$ or $\theta$) instead of $y$ written directly as a function of $x$.

Steps

Step 1: Recognise the parametric form

If you are given $x = f(\text{param})$ and $y = g(\text{param})$ instead of $y = h(x)$, do not try to eliminate the parameter first — it is often messy or impossible. Differentiate each equation separately with respect to the parameter instead.

Step 2: Differentiate $x$ and $y$ with respect to the parameter

Use the ordinary rules (product rule, chain rule, standard derivatives) to find $\dfrac{dx}{d\theta}$ (or $\dfrac{dx}{dt}$) and $\dfrac{dy}{d\theta}$ (or $\dfrac{dy}{dt}$).

Step 3: Divide — the parametric-derivative formula

This is justified by the chain rule: $\dfrac{dy}{d\theta} = \dfrac{dy}{dx}\cdot\dfrac{dx}{d\theta}$, so dividing recovers $\dfrac{dy}{dx}$.

Step 4: Simplify with trigonometric identities where possible …

Common Mistakes

Mistake 1: Forgetting the factor of $2$ from the multiple-angle terms.

Why it's wrong: $\frac{d}{d\theta}\cos2\theta=-2\sin2\theta$ and $\frac{d}{d\theta}\sin2\theta=2\cos2\theta$ — omitting the $2$ gives a derivative exactly half the correct size. Correct approach: apply the chain rule to every $2\theta$ term individually, don't just copy the single-angle derivative pattern.

Mistake 2: Sign error from the leading minus sign in $x=\cos\theta-\cos2\theta$. …