If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$, then find $\frac{dy}{dx}$ at $t = 2$.

> [!TLDR] > To find $\frac{dy}{dx}$ for parametric equations, we use the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. After calculating the derivatives and simplifying, we substitute the given value of $t$. The value of $\frac{dy}{dx}$ at $t=2$ is $\boxed{\frac{5}{3}}$. When variables $x$ and $y$ are both expressed in terms of a third variable, say $t$, these are called parametric equations. Here, $t$ is the parameter. To find $\frac{dy}{dx}$ in such a scenario, we cannot directly differentiate $y$ with respect to $x$ because $y$ is not given as an explicit function of $x$. Instead, we use the chain rule. Imagine we want to find the rate of change of $y$ with respect to $x$. We know how $y$ changes with $t$ (i.e., $\frac{dy}{dt}$) and how $x$ changes with $t$ (i.e., $\frac{dx}{dt}$). The chain rule allows us to link these rates: > [!FORMULA] > For parametric equations $x = f(t)$ and $y = g(t)$, the derivative $\frac{dy}{dx}$ is given by: > $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} $$ > provided that $\frac{dx}{dt} \neq 0$. This formula essentially says that the rate of change of $y$ with respect to $x$ is the ratio of how $y$ changes with $t$ to how $x$ changes with $t$. Let's apply this method step-by-step. 1. **Find $\frac{dx}{dt}$:** We are given $x = t + \frac{1}{t}$. We can rewrite $\frac{1}{t}$ as $t^{-1}$ to make differentiation easier using the power rule. $x = t + t^{-1}$ Differentiating $x$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(t) + \frac{d}{dt}(t^{-1})$ $\frac{dx}{dt} = 1 + (-1)t^{-1-1}$ $\frac{dx}{dt} = 1 - t^{-2}$ $\frac{dx}{dt} = 1 - \frac{1}{t^2}$ 2. **Find $\frac{dy}{dt}$:** We are given $y = t - \frac{1}{t}$. Similarly, rewrite $\frac{1}{t}$ as $t^{-1}$. $y = t - t^{-1}$ Differentiating $y$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt}(t) - \frac{d}{dt}(t^{-1})$ $\frac{dy}{dt} = 1 - (-1)t^{-1-1}$ $\frac{dy}{dt} = 1 + t^{-2}$ $\frac{dy}{dt} = 1 + \frac{1}{t^2}$ 3.

If x = t + (1)/(t) and y = t - (1)/(t), then find (dy)/ — CBSE Class 12 Mathematics 2026 PYQ

Question

x = t + \frac{1}{t}

CBSECBSE Class XII Board 2026Subjective· 2mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 $\frac{d^2y}{dx^2}=-\frac{1}{(1-\cos t)^2}.$ …

Start your 14-day free trial to unlock the full solution →

To find $\frac{dy}{dx}$ for parametric equations, we use the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . After calculating the derivatives and simplifying, we substitute the given value of $t$ . The value of $\frac{dy}{dx}$ at $t=2$ is $\boxed{\frac{5}{3}}$ .

When variables $x$ and $y$ are both expressed in terms of a third variable, say $t$ , these are called parametric equations. Here, $t$ is the parameter. To find $\frac{dy}{dx}$ in such a scenario, we cannot directly differentiate $y$ with respect to $x$ because $y$ is not given as an explicit function of $x$ .

Instead, we use the chain rule. Imagine we want to find the rate of change of $y$ with respect to $x$ . We know how $y$ changes with $t$ (i.e., $\frac{dy}{dt}$ ) and how $x$ changes with $t$ (i.e., $\frac{dx}{dt}$ ). The chain rule allows us to link these rates:

x = f(t)

This formula essentially says that the rate of change of $y$ with respect to $x$ is the ratio of how $y$ changes with $t$ to how $x$ changes with $t$ .

Let's apply this method step-by-step.

Find $\frac{dx}{dt}$ :

We are given $x = t + \frac{1}{t}$ . We can rewrite $\frac{1}{t}$ as $t^{-1}$ to make differentiation easier using the power rule.

$x = t + t^{-1}$

Differentiating $x$ with respect to $t$ :

$\frac{dx}{dt} = \frac{d}{dt}(t) + \frac{d}{dt}(t^{-1})$

$\frac{dx}{dt} = 1 + (-1)t^{-1-1}$

$\frac{dx}{dt} = 1 - t^{-2}$

$\frac{dx}{dt} = 1 - \frac{1}{t^2}$
Find $\frac{dy}{dt}$ :

We are given $y = t - \frac{1}{t}$ . Similarly, rewrite $\frac{1}{t}$ as $t^{-1}$ .

$y = t - t^{-1}$

Differentiating $y$ with respect to $t$ :

$\frac{dy}{dt} = \frac{d}{dt}(t) - \frac{d}{dt}(t^{-1})$

$\frac{dy}{dt} = 1 - (-1)t^{-1-1}$

$\frac{dy}{dt} = 1 + t^{-2}$

$\frac{dy}{dt} = 1 + \frac{1}{t^2}$
Calculate $\frac{dy}{dx}$ using the chain rule:

Now we use the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ .

$\frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}}$

To simplify this expression, find a common denominator in the numerator and denominator:

$\frac{dy}{dx} = \frac{\frac{t^2 + 1}{t^2}}{\frac{t^2 - 1}{t^2}}$

The $t^2$ terms cancel out:

$\frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1}$

Watch out

A common mistake is to forget to simplify the expression for $\frac{dy}{dx}$ before substituting the value of $t$ . While substituting early might sometimes work, simplifying first often prevents calculation errors and makes the substitution easier.
Evaluate $\frac{dy}{dx}$ at $t=2$ : …

Start your 14-day free trial to unlock the full solution →

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

CBSE 2026Set ANNUAL1 markMCQ
$x = a\cos\theta$
(a) $-\cot\theta$
(b) $\tan\theta$
(c) $a\tan\theta$
(d) None of these
›Reveal solutionSolution

For a parametric curve, $\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$ .

$x = a\cos\theta \Rightarrow \dfrac{dx}{d\theta} = -a\sin\theta$

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

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q.If x=a sec \theta and y=b tan \theta, then find \frac{dy}{dx}.$
›Reveal solutionSolution
$\dfrac{dy}{dx}=\dfrac{b}{a}\csc\theta$ .

Concept. For a curve given parametrically as $x=f(\theta),\ y=g(\theta)$ , $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$ .

Steps.

$x=a\sec\theta\Rightarrow \dfrac{dx}{d\theta}=a\sec\theta\tan\theta$ .

$y=b\tan\theta\Rightarrow \dfrac{dy}{d\theta}=b\sec^2\theta$ .

$\dfrac{dy}{dx}=\dfrac{b\sec^2\theta}{a\sec\theta\tan\theta}=\dfrac{b\sec\theta}{a\tan\theta}$ . …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$x = a \sec\theta$
(a) $\dfrac{b}{a} \sec\theta$
(b) $\dfrac{b}{a} \text{cosec}\,\theta$
(c) $\dfrac{b}{a} \cot\theta$
(d) none of these
›Reveal solutionSolution

Differentiate x and y separately with respect to the parameter θ, then divide.

$x = a\sec\theta \Rightarrow \dfrac{dx}{d\theta} = a\sec\theta\tan\theta$

$y = b\tan\theta \Rightarrow \dfrac{dy}{d\theta} = b\sec^2\theta$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Use parametric differentiation: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ .

Given $x = 2at^2$ , $y = at^4$ .

$\frac{dx}{dt} = 4at, \qquad \frac{dy}{dt} = 4at^3$

So: …

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$x = a\cos\theta$
(a) $\tan\theta$
(b) $-\cot\theta$
(c) $-\tan\theta$
(d) $\sec^2\theta$
›Reveal solutionSolution

Parametric differentiation gives $\frac{a\cos\theta}{-a\sin\theta}=-\cot\theta$ .

$\frac{dx}{d\theta}=-a\sin\theta$ , $\frac{dy}{d\theta}=a\cos\theta$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$x = at^2$
(a) $t$
(b) $\frac{1}{t}$
(c) $t^2$
(d) None of these
›Reveal solutionSolution

Parametric: $\frac{dy/dt}{dx/dt}=\frac{2a}{2at}=\frac1t$ .

$\frac{dx}{dt}=2at$ , $\frac{dy}{dt}=2a$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2018Set ANNUAL1 markMCQ
$x = t^{2}$
(a) $\dfrac{3}{2}$
(b) $\dfrac{3}{4t}$
(c) $\dfrac{3}{2t}$
(d) $\dfrac{3t}{2}$
›Reveal solutionSolution

For parametric curves, first find dy/dx as a function of t, then differentiate that again with respect to t and divide by dx/dt once more.

$x=t^2 \Rightarrow \dfrac{dx}{dt}=2t$ ; $y=t^3 \Rightarrow \dfrac{dy}{dt}=3t^2$

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3t^2}{2t} = \dfrac{3t}{2}$ …

Start your 14-day free trial to see the full solution →
CBSE 2016Set ANNUAL1 markMCQ
$x=t^2$
(a) $\dfrac{4t}{3}$
(b) $\dfrac{2t}{3}$
(c) $\dfrac{3}{2t}$
(d) $\dfrac{3}{4t}$
›Reveal solutionSolution

second derivative of a parametric curve

$x=t^2\Rightarrow \dfrac{dx}{dt}=2t$ ; $\ y=t^3\Rightarrow \dfrac{dy}{dt}=3t^2$ .

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

Differentiate again with respect to $x$ (using $\dfrac{d}{dx}=\dfrac{1}{dx/dt}\dfrac{d}{dt}$ ):

…

Start your 14-day free trial to see the full solution →

Q.If x=t+1tx = t + \frac{1}{t}x=t+t1​ and y=t−1ty = t - \frac{1}{t}y=t−t1​, then find dydx\frac{dy}{dx}dxdy​ at t=2t = 2t=2.

🎓Unlock everything free for 14 days

Q.If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$ , then find $\frac{dy}{dx}$ at $t = 2$ .

Unlock everything free for 14 days