(a) If $\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)$, then prove that $\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}$. OR (b) If $x = a\left(\cos\theta + \log\tan\frac{\theta}{2}\right)$ and $y = \sin\theta$, then find $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{4}$.

(a) If √(1 - x²) + √(1 - y²) = a(x - y), then prove tha — CBSE Class 12 Mathematics 2025 PYQ

Question

\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)

(OR)

(b) If

x = a\left(\cos\theta + \log\tan\frac{\theta}{2}\right)

and

y = \sin\theta

, then find

\frac{d^2y}{dx^2}

\theta = \frac{\pi}{4}

CBSECBSE Class XII Board 2025Subjective· 5mImportance★★★★★

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Part (a)Concept understanding — Implicit Differentiation

Implicit Differentiation

When $y$ isn't alone

You can differentiate $y=x^2+3x$ term by term because $y$ is written explicitly in terms of $x$ . But an equation like $x^2+y^2=25$ , or $x^3+y^3=6xy$ , does not give $y$ by itself — solving for $y$ is messy or downright impossible.

Implicit differentiation finds $\dfrac{dy}{dx}$ without isolating $y$ : treat $y$ as an unknown function of $x$ , differentiate the whole equation as it stands, then solve for $\dfrac{dy}{dx}$ .

The one key move: $y$ is really $y(x)$

Wherever $y$ appears, picture $y(x)$ hiding inside. Differentiating a $y$ -term therefore needs the chain rule, which tacks on a factor of $\dfrac{dy}{dx}$ :

\frac{d}{dx}\big(y^2\big)=2y\,\frac{dy}{dx}.

That extra $\dfrac{dy}{dx}$ on every $y$ -term is the whole trick.

The procedure

x

Worked example

For $x^2+y^2=25$ :

2x+2y\frac{dy}{dx}=0 \quad\Rightarrow\quad \frac{dy}{dx}=-\frac{x}{y}.

The answer naturally contains both $x$ and $y$ — that is normal here. To get the slope at a point on the curve, substitute the coordinates after differentiating; there is no need to solve for $y$ first. …

Part (b)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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Part (a): with $x=\sin\alpha,\ y=\sin\beta$ the relation forces $\alpha-\beta$ constant, giving $\dfrac{dy}{dx}=\sqrt{\dfrac{1-y^2}{1-x^2}}$ . Part (b): $\dfrac{dy}{dx}=\dfrac{\tan\theta}{a}$ and $\left.\dfrac{d^2y}{dx^2}\right|_{\pi/4}=\dfrac{2\sqrt2}{a^2}$ .

Part (a): prove $\dfrac{dy}{dx}=\sqrt{\dfrac{1-y^2}{1-x^2}}$

The terms $\sqrt{1-x^2}$ and $\sqrt{1-y^2}$ invite the substitution $x=\sin\alpha,\ y=\sin\beta$ with $\alpha,\beta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$ , so that $\sqrt{1-x^2}=\cos\alpha\ge0$ and $\sqrt{1-y^2}=\cos\beta\ge0$ . The equation becomes

\cos\alpha+\cos\beta=a(\sin\alpha-\sin\beta).

Apply sum-to-product identities:

2\cos\tfrac{\alpha+\beta}{2}\cos\tfrac{\alpha-\beta}{2}=a\cdot 2\cos\tfrac{\alpha+\beta}{2}\sin\tfrac{\alpha-\beta}{2}.

Cancelling $2\cos\tfrac{\alpha+\beta}{2}$ (nonzero in general) gives

\cot\tfrac{\alpha-\beta}{2}=a\ \Rightarrow\ \tfrac{\alpha-\beta}{2}=\text{const}\ \Rightarrow\ \alpha-\beta=\text{const}.

Differentiating $\alpha-\beta=$ const gives $d\alpha=d\beta$ . From $x=\sin\alpha$ , $dx=\cos\alpha\,d\alpha$ ; from $y=\sin\beta$ , $dy=\cos\beta\,d\beta=\cos\beta\,d\alpha$ . Therefore …

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

Showing the 12 most recent of 35 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$e^{-x} + e^{-y} = 2$
›Reveal solutionSolution
To find $\frac{dy}{dx}$ for an implicitly defined function, we differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ and applying the chain rule. The result is $\boxed{-e^{y-x}}$ .

When an equation relates $x$ and $y$ but does not explicitly express $y$ as a function of $x$ (like $y = f(x)$ ), we use a technique called implicit differentiation to find $\frac{dy}{dx}$ . The core idea is that even though $y$ isn't isolated, it is still a function of $x$ .

This means that when we differentiate a term involving $y$ with respect to $x$ , we must apply the chain rule. For example, if we differentiate $g(y)$ with respect to $x$ , we get $\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$ . This $\frac{dy}{dx}$ term is crucial and often the source of errors if overlooked.

Let's apply this to the given equation.

Differentiate both sides of the equation with respect to $x$ . The given equation is $e^{-x} + e^{-y} = 2$ . We apply the derivative operator $\frac{d}{dx}$ to every term:

$\frac{d}{dx}(e^{-x}) + \frac{d}{dx}(e^{-y}) = \frac{d}{dx}(2)$

Evaluate each derivative.

For the first term, $\frac{d}{dx}(e^{-x})$ :

Using the chain rule, if $u = -x$ , then $\frac{du}{dx} = -1$ .

So, $\frac{d}{dx}(e^{-x}) = e^{-x} \cdot \frac{d}{dx}(-x) = e^{-x} \cdot (-1) = -e^{-x}$ .

For the second term, $\frac{d}{dx}(e^{-y})$ :

This is where implicit differentiation comes in. We treat $y$ as a function of $x$ .

Using the chain rule, if $v = -y$ , then $\frac{dv}{dx} = \frac{d}{dx}(-y) = -1 \cdot \frac{dy}{dx}$ .

So, $\frac{d}{dx}(e^{-y}) = e^{-y} \cdot \frac{d}{dx}(-y) = e^{-y} \cdot \left(-\frac{dy}{dx}\right) = -e^{-y} \frac{dy}{dx}$ .

Watch out

A common mistake is to forget the $\frac{dy}{dx}$ term when differentiating expressions involving $y$ with respect to $x$ . Remember, $y$ is a function of $x$ .

For the right-hand side, $\frac{d}{dx}(2)$ : …
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CBSE 2026Set A1 markMCQ
$y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}}$
(a) $\frac{1}{2y-1}$
(b) $\frac{\cos x}{2y-1}$
(c) $\frac{\sin x}{2y-1}$
(d) $\frac{2y-1}{\cos x}$
›Reveal solutionSolution

$\frac{dy}{dx} = \frac{\cos x}{2y-1}$ .

The infinite nested radical satisfies $y = \sqrt{\sin x + y}$ , so

$y^2 = \sin x + y.$

Differentiate both sides implicitly with respect to $x$ :

$2y\frac{dy}{dx} = \cos x + \frac{dy}{dx}.$

Collect $\frac{dy}{dx}$ : …

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CBSE 2026Set A1 markMCQ
$x^n + y^n = a^n$
(a) $-\frac{x^{n-1}}{y^{n-1}}$
(b) $\frac{x^{n-1}}{y^{n-1}}$
(c) $-\frac{y^{n-1}}{x^{n-1}}$
(d) $nx^{n-1}$
›Reveal solutionSolution

$\frac{dy}{dx} = -\frac{x^{n-1}}{y^{n-1}}$ .

Differentiate $x^n + y^n = a^n$ implicitly ( $a$ constant):

$n x^{n-1} + n y^{n-1}\frac{dy}{dx} = 0.$

Solve: …

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CBSE 2026Set ANNUAL1 markMCQ
$2x + 3y = \sin y$
(a) $\frac{3}{\sin y - 2}$
(b) $\frac{2}{\cos y - 3}$
(c) $\frac{\cos y + 3}{2}$
(d) $\frac{2}{\cos y}$
›Reveal solutionSolution

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ , then solve for $dy/dx$ .

$2x+3y=\sin y$

Differentiating: $2 + 3\dfrac{dy}{dx} = \cos y\,\dfrac{dy}{dx}$

$2 = \dfrac{dy}{dx}(\cos y - 3)$

…

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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, $\dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta}$ .

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

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

…

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CBSE 2026Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Differentiate both sides of $2x+3y=\sin y$ with respect to $x$ (using the chain rule for the $y$ -terms), then collect $\dfrac{dy}{dx}$ on one side.

Given: $2x + 3y = \sin y$

Differentiate both sides w.r.t. $x$ :

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$

$2 + 3\frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$

Collect all $\dfrac{dy}{dx}$ terms on one side: …

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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.

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

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

$\dfrac{dy}{dx}=\dfrac{b\sec^2\theta}{a\sec\theta\tan\theta}=\dfrac{b\sec\theta}{a\tan\theta}$ . …
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CBSE 2025Set ANNUAL1 markMCQ
$x^2+y^2=2$
(a) $\dfrac{1-2x}{2y}$
(b) $\dfrac{2y}{1-2x}$
(c) $-\dfrac{x}{y}$
(d) $-\dfrac{y}{x}$
›Reveal solutionSolution

Differentiate $x^2+y^2=2$ implicitly with respect to $x$ , treating $y$ as a function of $x$ .

$\dfrac{d}{dx}(x^2+y^2)=\dfrac{d}{dx}(2)$

$2x+2y\dfrac{dy}{dx}=0$ …

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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.

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

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

…

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CBSE 2025Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ .

Start from $ax+by^2=\cos y$ and differentiate w.r.t. $x$ :

$\frac{d}{dx}(ax)+\frac{d}{dx}(by^2)=\frac{d}{dx}(\cos y)$

$a+2by\frac{dy}{dx}=-\sin y\frac{dy}{dx}.$

Gather the $\dfrac{dy}{dx}$ terms:

$2by\frac{dy}{dx}+\sin y\frac{dy}{dx}=-a$

$\frac{dy}{dx}\big(2by+\sin y\big)=-a.$

Hence

$\frac{dy}{dx}=\frac{-a}{2by+\sin y}.$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The value of dy/dx at (4, 1) of y³ - \sqrtx = 5 is ......................$
(a) 5/4
(b) 1/12
(c) 1/24
(d) 1/3
›Reveal solutionSolution

Differentiate the implicit relation $y^3 - \sqrt{x} = 5$ term by term with respect to $x$ , then substitute the point $(4, 1)$ .

Given: $y^3 - \sqrt{x} = 5$

Step 1 — differentiate implicitly w.r.t. $x$ :

$3y^2 \frac{dy}{dx} - \frac{1}{2\sqrt{x}} = 0$

Step 2 — solve for $dy/dx$ :

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}\cdot 3y^2} = \frac{1}{6y^2\sqrt{x}}$

…

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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: …

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Q.(a) If 1−x2+1−y2=a(x−y)\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)1−x2​+1−y2​=a(x−y), then prove that dydx=1−y21−x2\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}dxdy​=1−x21−y2​​.

Part (a): prove dydx=1−y21−x2\dfrac{dy}{dx}=\sqrt{\dfrac{1-y^2}{1-x^2}}dxdy​=1−x21−y2​​

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Q.(a) If $\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)$ , then prove that $\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}$ .

Part (a): prove $\dfrac{dy}{dx}=\sqrt{\dfrac{1-y^2}{1-x^2}}$

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