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) $\dfrac{3}{4}$

If x = t², y = t³, then (d² y)/(dx²) is (A) (3)/(2) (B) (3)/(4t) (C) (3)/(2t) (D) (3)/(4)

NCERT Exemplar · Q91

x = t^2

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Concept understanding — Parametric Second Derivative

Parametric Second Derivative

When a curve is given parametrically as $x = x(t)$ , $y = y(t)$ , its slope is

\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{y'(t)}{x'(t)},\qquad x'(t)\neq 0.

The second derivative $\dfrac{d^2y}{dx^2}$ measures how fast that slope changes — the concavity of the path. The catch is that $\frac{dy}{dx}$ comes out as a function of $t$ , but we need its rate of change with respect to $x$ .

The key idea

Differentiate the slope with respect to $t$ , then convert that $t$ -derivative into an $x$ -derivative by dividing by $dx/dt$ (chain rule):

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

Carrying this out with the quotient rule gives a compact closed form:

\frac{d^2y}{dx^2}=\frac{x'(t)\,y''(t)-y'(t)\,x''(t)}{\big[x'(t)\big]^3}.

Watch out

Do not write $\dfrac{d^2y}{dx^2}=\dfrac{d^2y/dt^2}{d^2x/dt^2}$ . The parametric second derivative is not the ratio of the second $t$ -derivatives — that tempting shortcut is wrong.

Worked illustration

For the cycloid $x=t-\sin t$ , $y=1-\cos t$ :

First derivatives: $\dfrac{dx}{dt}=1-\cos t$ , $\dfrac{dy}{dt}=\sin t$ , so $\dfrac{dy}{dx}=\dfrac{\sin t}{1-\cos t}$ .
Differentiate $\dfrac{dy}{dx}$ with respect to $t$ , then divide by $\dfrac{dx}{dt}=1-\cos t$ , which simplifies to $\frac{d^2y}{dx^2}=-\frac{1}{(1-\cos t)^2}.$ …

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For parametric equations, the second derivative is found by differentiating $\frac{dy}{dx}$ with respect to $t$ and dividing by $\frac{dx}{dt}$ . Here, $\frac{d^2 y}{dx^2} = \frac{3}{4t}$ , which is option (B).

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:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

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:

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

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

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

Now let’s apply it step by step.

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

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

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

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

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

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

\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{3}{2}t\right) = \frac{3}{2}

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

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$y=t^2+t^3$
›Reveal solutionSolution
For parametric curves, $\dfrac{d^2y}{dx^2}=\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)\Big/\dfrac{dx}{dt}$ ; evaluating at $t=1$ gives $-\dfrac43$ — (B).

Concept and Intuition

When $x,y$ are both given as functions of a parameter $t$ , we can't differentiate $dy/dx$ directly with respect to $x$ ; instead we differentiate the expression for $dy/dx$ (itself a function of $t$ ) with respect to $t$ , then divide by $dx/dt$ again to convert back to a derivative with respect to $x$ .

Step-by-Step Solution

$\dfrac{dy}{dt}=2t+3t^2$ , $\dfrac{dx}{dt}=1-4t^3$ .

$\dfrac{dy}{dx}=\dfrac{2t+3t^2}{1-4t^3}$ . Let $N=2t+3t^2$ , $D=1-4t^3$ , so $N'=2+6t$ , $D'=-12t^2$ .

$\dfrac{d}{dt}\left(\dfrac{N}{D}\right)=\dfrac{N'D-ND'}{D^2}$ . …
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AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$x=\sqrt2\, e^t(\sin t-\cos t)$
›Reveal solutionSolution
A parametric second derivative: compute $dy/dx=\cot t$ first (it collapses beautifully), then apply $\frac{d^2y}{dx^2}=\frac{d(dy/dx)/dt}{dx/dt}$ . Evaluated at $t=\pi/4$ , the answer is $-e^{-\pi/4}$ .

Concept and Intuition

For a parametric curve $x=x(t), y=y(t)$ , the first derivative is $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ , and the second derivative is not simply $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ — instead, you differentiate $\dfrac{dy}{dx}$ (a function of $t$ ) with respect to $t$ , then divide by $\dfrac{dx}{dt}$ again: $\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)=\dfrac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$ . Here the given $x,y$ have the special form of a rotating-and-expanding curve (logarithmic-spiral-like), and the ratio $dy/dt \div dx/dt$ simplifies neatly to $\cot t$ .

Step-by-Step Solution

$x=\sqrt2\,e^t(\sin t-\cos t)$ . Differentiate using product rule: $\dfrac{dx}{dt}=\sqrt2\left[e^t(\sin t-\cos t)+e^t(\cos t+\sin t)\right]=\sqrt2\,e^t\left[2\sin t\right]=2\sqrt2\,e^t\sin t$ .

$y=\sqrt2\,e^t(\sin t+\cos t)$ : $\dfrac{dy}{dt}=\sqrt2\left[e^t(\sin t+\cos t)+e^t(\cos t-\sin t)\right]=\sqrt2\,e^t\left[2\cos t\right]=2\sqrt2\,e^t\cos t$ .

$\dfrac{dy}{dx}=\dfrac{2\sqrt2\,e^t\cos t}{2\sqrt2\,e^t\sin t}=\cot t$ .

$\dfrac{d}{dt}(\cot t)=-\csc^2 t$ . …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$x = a\cos^3\theta$
›Reveal solutionSolution
This tests parametric differentiation (finding a second derivative $d^2y/dx^2$ for a curve given in parametric form).

Concept and Intuition

For parametric curves, $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$ , and the second derivative is $\dfrac{d^2y}{dx^2}=\dfrac{d(dy/dx)/d\theta}{dx/d\theta}$ — differentiate the first-derivative expression with respect to the parameter again, then divide by $dx/d\theta$ once more.

Step-by-Step Solution

$\dfrac{dx}{d\theta}=-3a\cos^2\theta\sin\theta$ , $\dfrac{dy}{d\theta}=3a\sin^2\theta\cos\theta$ .

$\dfrac{dy}{dx}=\dfrac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta}=-\dfrac{\sin\theta}{\cos\theta}=-\tan\theta$ .

Differentiate w.r.t. $\theta$ : $\dfrac{d}{d\theta}(-\tan\theta)=-\sec^2\theta$ .

$\dfrac{d^2y}{dx^2}=\dfrac{-\sec^2\theta}{dx/d\theta}=\dfrac{-\sec^2\theta}{-3a\cos^2\theta\sin\theta}=\dfrac{1}{3a\cos^4\theta\sin\theta}$ . …
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AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$x = f(\theta)$
›Reveal solutionSolution
This is the standard parametric second-derivative formula; carrying out the two differentiations gives $\dfrac{f'g''-g'f''}{(f')^3}$ .

Concept and Intuition

For a parametric curve, $\dfrac{dy}{dx}$ is itself a function of $\theta$ . To get $\dfrac{d^2y}{dx^2}$ we differentiate $\dfrac{dy}{dx}$ with respect to $\theta$ again, and then divide by $\dfrac{dx}{d\theta}$ once more (chain rule) — we do NOT simply differentiate twice with respect to $\theta$ directly.

Step-by-Step Solution

$\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta} = \dfrac{g'(\theta)}{f'(\theta)}$ .

$\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \dfrac{\dfrac{d}{d\theta}\left(\dfrac{g'(\theta)}{f'(\theta)}\right)}{f'(\theta)}$ .

Quotient rule on the inner derivative: $\dfrac{d}{d\theta}\left(\dfrac{g'(\theta)}{f'(\theta)}\right) = \dfrac{g''(\theta)f'(\theta)-g'(\theta)f''(\theta)}{(f'(\theta))^2}$ . …
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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$y=2t^2+3t-5$
›Reveal solutionSolution
A normal parallel to the X-axis means the tangent is vertical, i.e. $dx/dt=0$ while $dy/dt\ne0$ ; solving $dx/dt=0$ gives exactly two valid parameter values.

Concept and Intuition

For a parametric curve, the tangent direction is along $(dx/dt,\,dy/dt)$ and the normal is perpendicular to it. A normal parallel to the X-axis (i.e. horizontal) means the tangent itself is perpendicular to the X-axis, i.e. vertical — which happens exactly when $dx/dt=0$ (as long as $dy/dt\ne0$ there, so the curve doesn't also stall in $y$ ).

Step-by-Step Solution

Compute the derivatives with respect to the parameter $t$ :

$\frac{dx}{dt} = 3t^2-8t-3, \qquad \frac{dy}{dt}=4t+3$

For the normal to be parallel to the X-axis, we need the tangent vertical, i.e.

$\frac{dx}{dt}=0$

Solve $3t^2-8t-3=0$ :

$t = \frac{8\pm\sqrt{64+4\cdot3\cdot3}}{2\cdot3} = \frac{8\pm\sqrt{100}}{6} = \frac{8\pm10}{6}$ $t=3 \quad\text{or}\quad t=-\frac13$

Check $dy/dt\ne0$ at these points (needed so the tangent is genuinely vertical, not a cusp/degenerate point):

At $t=3$ : $dy/dt = 4(3)+3=15\ne0$ . Valid. …
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AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
$x = a[\cos\theta + \log(\tan(\theta/2))]$
›Reveal solutionSolution
Differentiating both $x$ and $y$ with respect to the parameter $\theta$ and simplifying using the double-angle identity gives $\dfrac{dy}{dx}=\tan\theta$ .

Concept and Intuition

For parametric curves, $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$ . The log term in $x(\theta)$ differentiates using the chain rule and simplifies neatly with the identity $2\sin(\theta/2)\cos(\theta/2)=\sin\theta$ .

Step-by-Step Solution

$y=a\sin\theta \Rightarrow \dfrac{dy}{d\theta}=a\cos\theta$ .

$x=a\left[\cos\theta+\log\left(\tan\dfrac{\theta}{2}\right)\right]$ .

$\dfrac{dx}{d\theta}=a\left[-\sin\theta+\dfrac{1}{\tan(\theta/2)}\cdot\sec^2\left(\dfrac\theta2\right)\cdot\dfrac12\right]$ .

Simplify the log-derivative term: $\dfrac{\cos(\theta/2)}{\sin(\theta/2)}\cdot\dfrac{1}{2\cos^2(\theta/2)}=\dfrac{1}{2\sin(\theta/2)\cos(\theta/2)}=\dfrac{1}{\sin\theta}$ .

So $\dfrac{dx}{d\theta}=a\left[-\sin\theta+\dfrac{1}{\sin\theta}\right]=a\cdot\dfrac{1-\sin^2\theta}{\sin\theta}=a\cdot\dfrac{\cos^2\theta}{\sin\theta}$ . …
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AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
$y = 4\cos^3(t)$
›Reveal solutionSolution
Differentiating both parametric equations with respect to $t$ and simplifying the ratio gives $\dfrac{dy}{dx}=-\cot t$ .

Concept and Intuition

For parametric curves, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ ; applying the chain rule to the cube powers of trig functions and cancelling common factors gives a clean trig ratio.

Step-by-Step Solution

$y=4\cos^3t \Rightarrow \dfrac{dy}{dt}=4\cdot3\cos^2t\cdot(-\sin t)=-12\cos^2t\sin t$ .

$x=4\sin^3t \Rightarrow \dfrac{dx}{dt}=4\cdot3\sin^2t\cdot\cos t=12\sin^2t\cos t$ .

$\dfrac{dy}{dx}=\dfrac{-12\cos^2t\sin t}{12\sin^2t\cos t}=-\dfrac{\cos t}{\sin t}=-\cot t$ . …
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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 understanding — Parametric Second Derivative

Parametric Second Derivative

The key idea

Worked illustration

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

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.If x=t2x = t^2x=t2, y=t3y = t^3y=t3, then d2ydx2\dfrac{d^2 y}{dx^2}dx2d2y​ is (A) 32\dfrac{3}{2}23​ (B) 34t\dfrac{3}{4t}4t3​ (C) 32t\dfrac{3}{2t}2t3​ (D) 34\dfrac{3}{4}43​

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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) $\dfrac{3}{4}$

Unlock everything free for 14 days