Q.Find $\frac{dy}{dx}$ in the following: $x = a \left(\cos t + \log \tan \frac{t}{2}\right), y = a \sin t$

The key idea is Implicit Differentiation using parametric equations — we find $\frac{dy}{dx}$ as $\frac{dy/dt}{dx/dt}$.

First, differentiate $y$ with respect to $t$:

Next, differentiate $x$ with respect to $t$. The derivative of $a \cos t$ is $-a \sin t$. For $a \log \tan \frac{t}{2}$, use the chain rule:

Thus, …

When a curve is given in parametric form — $x = f(t)$, $y = g(t)$ — the derivative $\frac{dy}{dx}$ is not directly available. But the chain rule gives a clean path:

This is the core idea of parametric differentiation. Instead of eliminating $t$ (which is often messy), we differentiate each equation with respect to the parameter $t$ and then take the ratio.

Let’s apply this to the given equations.

1. Differentiate $y = a \sin t$ with respect to $t$. Since $a$ is a constant,

2. Differentiate $x = a \left( \cos t + \log \tan \frac{t}{2} \right)$ with respect to $t$. Factor out the constant $a$:

The derivative of $\cos t$ is $-\sin t$.

For the logarithmic term, use the chain rule:

Now $\frac{d}{dt}\left[ \tan \frac{t}{2} \right] = \sec^2 \frac{t}{2} \cdot \frac{1}{2}$ (by the chain rule again, since derivative of $\frac{t}{2}$ is $\frac{1}{2}$).

So

Recall that $\frac{1}{\tan \theta} = \cot \theta$ and $\sec^2 \theta = 1 + \tan^2 \theta$, but here a simpler simplification is:

And $\sin \frac{t}{2} \cos \frac{t}{2} = \frac{1}{2} \sin t$ (using the double-angle identity $\sin 2\theta = 2 \sin \theta \cos \theta$).

Therefore,

Including the factor $\frac{1}{2}$ from earlier:

$\frac{d}{dt} \left[ \log \tan \frac{t}{2} \right] = \frac{1}{2} \cdot \frac{2}{\sin t} = \frac{1}{\sin t}.$ …

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: Dropping the $\frac{1}{2}$ factor when differentiating $\tan\frac{t}{2}$.

Why it's wrong: the chain rule on $\tan\frac{t}{2}$ contributes $\sec^2\frac{t}{2}\cdot\frac{1}{2}$, and forgetting the $\frac{1}{2}$ doubles that part of the answer. Correct approach: differentiate the inner argument $\frac{t}{2}$ explicitly as its own step.

Mistake 2: Not converting $\frac{1}{\sin\frac{t}{2}\cos\frac{t}{2}}$ to $\frac{2}{\sin t}$ via the double-angle identity. …