(a) If $y = \left[\log\left(\sqrt{x} + \dfrac{1}{\sqrt{x}}\right)\right]^2$, then show that $x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2$. OR (b) If $x\sqrt{1 + y} + y\sqrt{1 + x} = 0$, $-1 < x < 1$, $x \ne y$, then prove that $\dfrac{dy}{dx} = -\dfrac{1}{(1 + x)^2}$.

(a) If y = [log(√(x) + dfrac1√(x))]², then show that x(x + 1)² y₂ + (x + 1)² y₁ = 2. OR (b) If x√(1 + y) + y√(

Question

y = \left[\log\left(\sqrt{x} + \dfrac{1}{\sqrt{x}}\right)\right]^2

(OR)

(b) If

x\sqrt{1 + y} + y\sqrt{1 + x} = 0

-1 < x < 1

x \ne y

, then prove that

\dfrac{dy}{dx} = -\dfrac{1}{(1 + x)^2}

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★

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Part (a)Concept understanding — Successive Differentiation

Successive Differentiation — Repeated Slopes

Differentiate a function, then differentiate the result, then differentiate that, and so on. Each pass produces a new function describing a deeper layer of change. The physics picture makes it concrete: position $\to$ velocity (1st derivative) $\to$ acceleration (2nd) $\to$ jerk (3rd). Every step asks the same question: "how does the previous rate of change itself change?"

The definition and notation

If $y=f(x)$ , its successive derivatives are written $y_1,y_2,\dots,y_n$ , equivalently $f'(x),f''(x),\dots,f^{(n)}(x)$ or $\dfrac{dy}{dx},\dfrac{d^2y}{dx^2},\dots,\dfrac{d^ny}{dx^n}$ . Each one is the derivative of the previous:

\frac{d^ny}{dx^n}=\frac{d}{dx}\!\left(\frac{d^{n-1}y}{dx^{n-1}}\right).

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

Seeing the pattern

Take $y=x^4$ : $\;y_1=4x^3,\;y_2=12x^2,\;y_3=24x,\;y_4=24,\;y_5=0$ . Each differentiation drops the degree by one, so a degree- $n$ polynomial has a constant $n$ th derivative and vanishing higher ones. Other families behave differently: $e^{ax}$ never dies out ( $\frac{d^n}{dx^n}e^{ax}=a^n e^{ax}$ ), and $\sin x$ cycles every four steps ( $\sin\to\cos\to-\sin\to-\cos$ ).

Tip

Power rule applied $n$ times: $\dfrac{d^n}{dx^n}(x^m)=m(m-1)\cdots(m-n+1)\,x^{m-n}$ for $n\le m$ .

Watch out

$\dfrac{d^2y}{dx^2}$ is not $\left(\dfrac{dy}{dx}\right)^2$ — a second derivative is not the square of the first derivative. …

Part (b)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. …

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Part (a): by successive differentiation, $y=\big[\log(x+\sqrt{1+x^2})\big]^2$ satisfies $(1+x^2)y_2+xy_1=2$ . Part (b): simplifying the implicit relation to $y=-\dfrac{x}{1+x}$ gives $\dfrac{dy}{dx}=-\dfrac{1}{(1+x)^2}$ .

Part (a): successive differentiation

Note

The identity $x(x+1)^2y_2+(x+1)^2y_1=2$ does not hold for $y=\big[\log(\sqrt x+\tfrac{1}{\sqrt x})\big]^2$ ; that expression is a transcription of the standard problem $y=\big[\log\!\big(x+\sqrt{1+x^2}\big)\big]^2$ , for which the correct second-order relation is $(1+x^2)y_2+xy_1=2$ . The intended technique — successive differentiation — is shown below on the standard form.

Let $y=\big[\log(x+\sqrt{1+x^2})\big]^2$ . Recall $\dfrac{d}{dx}\log(x+\sqrt{1+x^2})=\dfrac{1}{\sqrt{1+x^2}}$ . By the chain rule,

y_1=2\log(x+\sqrt{1+x^2})\cdot\frac{1}{\sqrt{1+x^2}}=\frac{2\log(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}.

Clearing the radical:

\sqrt{1+x^2}\,y_1=2\log(x+\sqrt{1+x^2}).

Differentiate both sides once more. On the left use the product rule with $\dfrac{d}{dx}\sqrt{1+x^2}=\dfrac{x}{\sqrt{1+x^2}}$ ; on the right the derivative is $\dfrac{2}{\sqrt{1+x^2}}$ :

\sqrt{1+x^2}\,y_2+\frac{x}{\sqrt{1+x^2}}\,y_1=\frac{2}{\sqrt{1+x^2}}.

Multiply throughout by $\sqrt{1+x^2}$ : …

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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 49 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
$\frac{d^2}{dx^2}(\sin 2x) =$
(a) $4\sin 2x$
(b) $4\cos^2 2x$
(c) $-4\sin 2x$
(d) $2\sin 4x$
›Reveal solutionSolution

$\frac{d^2}{dx^2}(\sin 2x) = -4\sin 2x$ .

First derivative (chain rule):

$\frac{d}{dx}(\sin 2x) = 2\cos 2x.$

Second derivative: …

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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 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
$Q. If y = 8e⁻³ˣ, find d²y/dx².$
›Reveal solutionSolution

Differentiate $y=8e^{-3x}$ twice using the chain rule.

$\dfrac{dy}{dx} = 8\cdot(-3)e^{-3x} = -24e^{-3x}$

…

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CBSE 2026Set ANNUAL1 mark
$Q.Find the second derivative for the function y=sin x + e^{2x}.$
›Reveal solutionSolution
$y''=-\sin x+4e^{2x}$ .

Concept. The second derivative is found by differentiating the first derivative; use $\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{d}{dx}e^{kx}=k\,e^{kx}$ .

Steps.

$y=\sin x+e^{2x}$ .

First derivative: $y'=\cos x+2e^{2x}$ . …
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CBSE 2025Set ANNUAL1 mark
$y = \log x$
›Reveal solutionSolution

Differentiate $y=\log x$ twice.

$y=\log x \implies \dfrac{dy}{dx}=\dfrac1x$

…

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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
$y = 2\sin x + 3\cos x$
(a) $y$
(b) $\dfrac{1}{y}$
(c) $-y$
(d) $-\dfrac{1}{y}$
›Reveal solutionSolution

Differentiate twice — the second derivative comes back around to -y, a classic SHM-type result.

$y = 2\sin x + 3\cos x$

$\dfrac{dy}{dx} = 2\cos x - 3\sin x$

…

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CBSE 2025Set ANNUAL1 mark
$x\cos x$
›Reveal solutionSolution

Apply the product rule twice.

Let $y=x\cos x$ .

First derivative (product rule on $x$ and $\cos x$ ):

$y'=\frac{d}{dx}(x)\cos x+x\frac{d}{dx}(\cos x)=1\cdot\cos x+x(-\sin x)=\cos x-x\sin x.$

Second derivative (differentiate $\cos x$ and the product $x\sin x$ ): …

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Q.(a) If $y = \left[\log\left(\sqrt{x} + \dfrac{1}{\sqrt{x}}\right)\right]^2$ , then show that $x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2$ .

Part (a)Concept understanding — Successive Differentiation

Successive Differentiation — Repeated Slopes

The definition and notation

Seeing the pattern

Part (b)Concept understanding — Implicit Differentiation

Implicit Differentiation

When $y$ isn't alone

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

The procedure

Worked example

Part (a) (successive differentiation)

Part (a): successive differentiation

Q.(a) If y=[log⁡(x+1x)]2y = \left[\log\left(\sqrt{x} + \dfrac{1}{\sqrt{x}}\right)\right]^2y=[log(x​+x​1​)]2, then show that x(x+1)2y2+(x+1)2y1=2x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2x(x+1)2y2​+(x+1)2y1​=2.

Part (a): successive differentiation

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Q.(a) If $y = \left[\log\left(\sqrt{x} + \dfrac{1}{\sqrt{x}}\right)\right]^2$ , then show that $x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2$ .

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