Q.Find $\frac{dy}{dx}$ in the following: $x = 4t, y = \frac{4}{t}$

Idea: $x$ and $y$ are both given through a parameter $t$, so use $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ (parametric differentiation).

Differentiate w.r.t. $t$:

$\frac{dx}{dt} = 4, \qquad \frac{dy}{dt} = \frac{d}{dt}\left(4t^{-1}\right) = -\frac{4}{t^2}.$ …

Here the curve is not written as $y$ in terms of $x$ directly; both coordinates depend on a parameter $t$. As $t$ changes, $x$ and $y$ each change, and the slope $\frac{dy}{dx}$ is the ratio of their rates of change with respect to $t$.

Step 1 — differentiate each with respect to $t$

Step 2 — take the ratio

Step 3 — optional forms in $x$ or $y$ …

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: Mishandling the negative exponent when differentiating $4/t$.

Why it's wrong: writing $\frac{d}{dt}(4t^{-1}) = 4t^{-2}$ (dropping the sign) instead of $-4t^{-2}$ is a very common slip with negative powers. Correct approach: recall $\frac{d}{dt}t^{n}=nt^{n-1}$ applies for negative $n$ too, so $n=-1$ gives $-t^{-2}$.

Mistake 2: Leaving the answer in a form that mixes $t$ and unsimplified fractions. …