Q.Find $\frac{dy}{dx}$ in the following: $x = a \sec \theta, y = b \tan \theta$

Concept: Parametric Differentiation — when both $x$ and $y$ are given in terms of a parameter $\theta$, we use $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.

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

Now apply the chain rule for parametric equations:

Simplify by cancelling one $\sec\theta$: …

When a curve is given in parametric form — both $x$ and $y$ expressed in terms of a third variable (here $\theta$) — we cannot directly write $y$ as a function of $x$. Instead, we find $\frac{dy}{dx}$ by dividing the derivative of $y$ with respect to the parameter by the derivative of $x$ with respect to the parameter. This works because of the chain rule: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$, provided $dx/d\theta \neq 0$.

Let’s apply this to the given equations.

1. Differentiate $x$ with respect to $\theta$. $x = a \sec \theta$. The derivative of $\sec \theta$ is $\sec \theta \tan \theta$. So

2. Differentiate $y$ with respect to $\theta$. $y = b \tan \theta$. The derivative of $\tan \theta$ is $\sec^2 \theta$. So

3. Form the ratio $\frac{dy}{dx}$. Using the parametric formula:

4. Simplify the expression. Cancel one factor of $\sec \theta$:

Now recall that $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$. Substituting:

And $\frac{1}{\sin \theta} = \csc \theta$, so

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

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: Confusing $\frac{d}{d\theta}\sec\theta$ with $\sec^2\theta$.

Why it's wrong: $\sec^2\theta$ is the derivative of $\tan\theta$, not of $\sec\theta$ — mixing them up is one of the most common trig-derivative slips. Correct approach: keep the two standard derivatives ($\sec\theta\tan\theta$ and $\sec^2\theta$) clearly separated and double-check which function is which.

Mistake 2: Leaving the answer as an un-simplified ratio of secants and tangents. …