1 markMCQSection A

If $x = t^2$ and $y = t^3$ , then $\frac{d^2y}{dx^2}$ is equal to :

Q: If $x = t^2$ and $y = t^3$, then $\frac{d^2y}{dx^2}$ is equal to :

The correct option is **B) $\frac{3}{4t}$**. **Working:** 1. $dx/dt = 2t$, $dy/dt = 3t^2$. 2. $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = 3t^2/(2t) = \frac{3}{2}t$. 3. $\frac{d^2y}{dx^2} = \frac{d}{dx}[\frac{3}{2}t] = \frac{d}{dt}[\frac{3}{2}t] \div (dx/dt) = \frac{3}{2} \div (2t) = \frac{3}{4t}$.

Q: Why divide by $dx/dt$ again for the second derivative?

Because $\frac{d^2y}{dx^2} = \frac{d}{dx}(dy/dx)$, and to convert d/dx into $\frac{d}{dt}$ in parametric form we divide by $dx/dt$. Skipping this step is the most common error.

Q: Is $\frac{d^2y}{dx^2}$ the same as differentiating $\frac{dy}{dx}$ twice with respect to t?

No. It is $\frac{d}{dt}(dy/dx)$ divided by $dx/dt$. Differentiating twice with respect to t would give a different, incorrect expression.

Calculus (Differentiation)

Parametric Differentiation — second derivative

Options

(A)

\frac{3}{2}

(B)

\frac{3}{4t}

(C)

\frac{1}{2t^2}

(D)

\frac{3}{2t}

Official Answer

The correct option is B) $\frac{3}{4t}$ .

Working:

$dx/dt = 2t$ , $dy/dt = 3t^2$ .
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = 3t^2/(2t) = \frac{3}{2}t$ .
$\frac{d^2y}{dx^2} = \frac{d}{dx}[\frac{3}{2}t] = \frac{d}{dt}[\frac{3}{2}t] \div (dx/dt) = \frac{3}{2} \div (2t) = \frac{3}{4t}$ .

parametric differentiationsecond order derivativedy/dx = (3/2)tdivide by dx/dt3/(4t)chain rulex = t^2 y = t^3

Marking Scheme

11 mark: correct $\frac{dy}{dx} = \frac{3}{2}t$ and $\frac{d^2y}{dx^2} = \frac{3}{4t}$ via division by $dx/dt$ (option B).

Hint

Find $\frac{dy}{dx} = \frac{3}{2}t$ first, then $\frac{d^2y}{dx^2} = [\frac{d}{dt}(dy/dx)] \div (dx/dt)$ ; do not forget the second division by $2t$ .

Quick Oral Answer

$\frac{dy}{dx}$ is $\frac{3}{2}t$ ; differentiating with respect to t gives 3/2, and dividing again by $dx/dt = 2t$ gives $\frac{d^2y}{dx^2} = \frac{3}{4t}$ .

Analysis & Explanation

This tests the second-order derivative of a parametric function — a common exam trap.

Concept:

For parametric $x(t)$ , $y(t)$ : $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ .
Crucially, $\frac{d^2y}{dx^2} = \frac{d}{dt}(dy/dx) \div (dx/dt)$ — you must divide again by $dx/dt$ , NOT differentiate $dy/dx$ with respect to t alone.

Why B is correct:

$\frac{dy}{dx} = \frac{3}{2}t$ ; differentiating w.r.t. t gives 3/2; dividing by $dx/dt = 2t$ gives $\frac{3}{4t}$ .

Why the distractors are wrong:

A (3/2): stops at $\frac{d}{dt}(dy/dx)$ and forgets to divide by $dx/dt = 2t$ .
C ( $\frac{1}{2t^2}$ ) and D ( $\frac{3}{2t}$ ): come from mis-dividing or omitting the factor 2 from dx/dt.

Common Mistakes

1Forgetting to divide the second time by $dx/dt$ , giving the wrong answer $\frac{3}{2}$ .
2Differentiating $\frac{dy}{dx}$ with respect to x directly, which is not possible in parametric form.
3Arithmetic slip: $\frac{3}{2} \div (2t)$ is $\frac{3}{4t}$ , not $\frac{3}{2t}$ .

Interesting Facts

The curve $x = t^2$ , $y = t^3$ is a semicubical parabola ( $y^2 = x^3$ ), famous as the first curve whose arc length was computed algebraically (by William Neile, 1657).

The recurring rule $\frac{d^2y}{dx^2} = [\frac{d}{dt}(dy/dx)] \div (dx/dt)$ is the single most tested trap in parametric calculus MCQs.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Why divide by $dx/dt$ again for the second derivative?

Because $\frac{d^2y}{dx^2} = \frac{d}{dx}(dy/dx)$ , and to convert d/dx into $\frac{d}{dt}$ in parametric form we divide by $dx/dt$ . Skipping this step is the most common error.

Is $\frac{d^2y}{dx^2}$ the same as differentiating $\frac{dy}{dx}$ twice with respect to t?

No. It is $\frac{d}{dt}(dy/dx)$ divided by $dx/dt$ . Differentiating twice with respect to t would give a different, incorrect expression.