Q.If $x = t^2$, $y = t^3$, then $\dfrac{d^2 y}{dx^2}$ is
(A) $\dfrac{3}{2}$
(B) $\dfrac{3}{4t}$
(C) $\dfrac{3}{2t}$
(D) $\dfrac{3}{4}$

Concept: Parametric Second Derivative — we differentiate $y$ with respect to $x$ by first finding $\frac{dy}{dx}$ as $\frac{dy/dt}{dx/dt}$, then differentiate that result with respect to $x$ using $\frac{d}{dx} = \frac{d/dt}{dx/dt}$.

Step 1: Compute first derivatives.

$\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 3t^2$.

Step 2: First derivative:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3t}{2}$.

Step 3: Second derivative: …

When a curve is given in parametric form — $x = f(t)$, $y = g(t)$ — the first derivative $\frac{dy}{dx}$ is obtained by the chain rule:

But the second derivative $\frac{d^2 y}{dx^2}$ is not simply $\frac{d^2 y/dt^2}{d^2 x/dt^2}$. That’s a common mistake. Instead, think of $\frac{dy}{dx}$ as a function of $t$, and then differentiate it with respect to $x$ using the chain rule again:

Since $\frac{dt}{dx} = 1 / \frac{dx}{dt}$, the formula becomes:

Now let’s apply it step by step.

1. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ Given $x = t^2$ and $y = t^3$:

2. Compute the first derivative $\frac{dy}{dx}$

Notice this is a simple linear function of $t$.

3. Differentiate $\frac{dy}{dx}$ with respect to $t$

4. Apply the formula for $\frac{d^2 y}{dx^2}$ …

Method: Finding $\dfrac{d^2y}{dx^2}$ for a Parametric Curve

The two-derivative trap in parametric problems is thinking you can just take second $t$-derivatives of $x$ and $y$ separately and divide — you can't. Second derivatives with respect to $x$ must be built up in two deliberate stages through $t$.

Steps

Step 1: Find the first derivative in terms of $t$

This gives $\dfrac{dy}{dx}$ as some expression in $t$ — call it $g(t)$.

Step 2: Differentiate that expression with respect to $t$ (not $x$)

Compute $\dfrac{d}{dt}\big(g(t)\big)$ using ordinary differentiation rules in $t$.

Step 3: Convert the $t$-derivative into an $x$-derivative by dividing by $dx/dt$

$\frac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$ …

Common Mistakes

Mistake 1: Computing $\dfrac{d^2y}{dx^2}$ as $\dfrac{d^2y/dt^2}{d^2x/dt^2}$

Why it's wrong: the chain rule for a second derivative does not work like the first-derivative ratio — squaring the process doesn't square the formula. This shortcut silently gives a completely different (and wrong) expression. Correct approach: always go through the two-stage process — first derivative as a $t$-ratio, then differentiate that ratio with respect to $t$ and divide by $dx/dt$ once more.

Mistake 2: Forgetting to divide by $dx/dt$ after differentiating $dy/dx$ with respect to $t$ …