Q.Find $\frac{dy}{dx}$ in the following: $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$

Idea: parametric differentiation — find $\frac{dx}{dt}$ and $\frac{dy}{dt}$, then $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$. Use $\frac{d}{dt}(\cos 2t)^{-1/2}=\frac{\sin 2t}{(\cos 2t)^{3/2}}$.

Differentiate $x=\sin^3 t\,(\cos 2t)^{-1/2}$:

using $\sin 2t=2\sin t\cos t$, $\cos 2t=\cos^2 t-\sin^2 t$.

Differentiate $y=\cos^3 t\,(\cos 2t)^{-1/2}$:

Divide: …

Both $x$ and $y$ are functions of the parameter $t$, so we differentiate each with respect to $t$ and take the ratio. Write $x=\sin^3 t\,(\cos 2t)^{-1/2}$ and $y=\cos^3 t\,(\cos 2t)^{-1/2}$, and recall $\frac{d}{dt}\cos 2t=-2\sin 2t$, so $\frac{d}{dt}(\cos 2t)^{-1/2}=\frac{\sin 2t}{(\cos 2t)^{3/2}}$.

Step 1 — differentiate $x$

By the product rule,

Factor $(\cos 2t)^{-3/2}$ and use $\sin 2t=2\sin t\cos t$, $\cos 2t=\cos^2 t-\sin^2 t$:

Step 2 — differentiate $y$

Similarly,

Step 3 — divide

The $(\cos 2t)^{3/2}$ factors cancel: …

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 chain rule on $(\cos2t)^{-1/2}$.

Why it's wrong: $\frac{d}{dt}(\cos2t)^{-1/2} = \frac{\sin2t}{(\cos2t)^{3/2}}$ (from the power rule times $\frac{d}{dt}\cos2t=-2\sin2t$) — dropping this term entirely misses half of each product-rule expansion. Correct approach: treat $(\cos2t)^{-1/2}$ as its own composite function and differentiate it explicitly before combining with the product rule.

Mistake 2: Dividing $dy/dt$ by $dx/dt$ before cancelling the shared $(\cos2t)^{3/2}$ factor. …