If $y = \sqrt{ax + b}$, prove that $y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$. OR If $f(x) = \begin{cases} ax + b, & 0 < x \leq 1 \\ 2x^2 - x, & 1 < x < 2 \end{cases}$ is a differentiable function in $(0, 2)$, then find the values of $a$ and $b$.

If y = √(ax + b), prove that y(d²y)/(dx²) + ((dy)/(dx)) — CBSE Class 12 Mathematics 2023 PYQ

Question

y = \sqrt{ax + b}

(OR)

f(x) = \begin{cases} ax + b, & 0 < x \leq 1 \\ 2x^2 - x, & 1 < x < 2 \end{cases}

CBSECBSE Class XII Board 2023Subjective· 2mImportance★★★★★

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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. …

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For $y = \sqrt{ax + b}$ , compute the first and second derivatives, substitute into the given expression, and simplify using the relationship $y^2 = ax + b$ to show the identity holds.

The key insight here is that $y$ and its derivatives are all algebraically related through the original equation $y^2 = ax + b$ . When we differentiate a square root function twice, the second derivative introduces a factor that exactly cancels with the square of the first derivative when weighted by $y$ .

Let's work through the derivatives systematically and see how the algebra unfolds.

Finding the first derivative

Starting with $y = \sqrt{ax + b} = (ax + b)^{1/2}$ , we apply the chain rule:

\frac{dy}{dx} = \frac{1}{2}(ax + b)^{-1/2} \cdot a = \frac{a}{2\sqrt{ax + b}} = \frac{a}{2y}

This is our first key relationship: $\frac{dy}{dx} = \frac{a}{2y}$ .

Finding the second derivative

Now differentiate $\frac{dy}{dx} = \frac{a}{2y}$ with respect to $x$ . Since $y$ depends on $x$ , we use the quotient rule (or chain rule):

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{a}{2y}\right) = \frac{a}{2} \cdot \frac{d}{dx}(y^{-1}) = \frac{a}{2} \cdot (-1)y^{-2} \cdot \frac{dy}{dx}

= -\frac{a}{2y^2} \cdot \frac{dy}{dx}

Substituting $\frac{dy}{dx} = \frac{a}{2y}$ :

\frac{d^2y}{dx^2} = -\frac{a}{2y^2} \cdot \frac{a}{2y} = -\frac{a^2}{4y^3}

Verifying the identity

Now we substitute both derivatives into the left-hand side of the expression we need to prove:

y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = y \cdot \left(-\frac{a^2}{4y^3}\right) + \left(\frac{a}{2y}\right)^2

= -\frac{a^2}{4y^2} + \frac{a^2}{4y^2}

$= 0$ …

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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 22 on this concept.

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 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
$y = 2\sin x + 3\cos x$
(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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CBSE 2024Set D1 markMCQ
$y = x^{20}$
(a) $x^{18}$
(b) $20x^{19}$
(c) $380x^{18}$
(d) $x^{19}$
›Reveal solutionSolution

$\frac{d^2y}{dx^2} = 380x^{18}$ .

Use the power rule $\frac{d}{dx}x^n = nx^{n-1}$ twice.

First derivative:

$\frac{dy}{dx} = 20x^{19}.$

Second derivative: …

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CBSE 2024Set ANNUAL1 markMCQ
$y=\log x$
(a) $-\frac{1}{x^2}$
(b) $\frac{1}{x^2}$
(c) $-\frac{1}{x}$
(d) $\frac{1}{x}$
›Reveal solutionSolution

Differentiate y=log x twice: first derivative is 1/x, second derivative is -1/x^2.

$y=\log x \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x}=x^{-1}$

…

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CBSE 2024Set ANNUAL1 mark
$\log x$
›Reveal solutionSolution

Differentiate $\log x$ twice.

Let $y = \log x$ . The first derivative is

$\dfrac{dy}{dx} = \dfrac{1}{x} = x^{-1}.$

Differentiating again, …

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CBSE 2023Set E1 markMCQ
$\dfrac{d^{2}}{dx^{2}}\left(e^{5x}\right) =$
(a) $e^{5x}$
(b) $10e^{5x}$
(c) $5e^{5x}$
(d) $25e^{5x}$
›Reveal solutionSolution

$\dfrac{d}{dx}e^{5x}=5e^{5x}$ , and differentiating again gives $25e^{5x}$ .

First derivative: $\dfrac{d}{dx}e^{5x}=5e^{5x}$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$y=x \cdot \log_e x$
(a) $\frac{1}{1+x}$
(b) $\frac{1}{x}$
(c) $\log_e(1+x)$
(d) $1+\log_e x$
›Reveal solutionSolution

Differentiate $y=x\log_e x$ twice using the product rule.

Given $y=x\log_e x$ .

First derivative (product rule, $u=x$ , $v=\log_e x$ ):

$\dfrac{dy}{dx}=1\cdot\log_e x + x\cdot\dfrac{1}{x}=\log_e x+1$

…

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CBSE 2022Set ANNUAL1 markMCQ
$\dfrac{d^2}{dx^2}\sin(3x+5) = ?$
(a) $\sin(3x+5)$
(b) $9\cos(3x+5)$
(c) $-9\sin(3x+5)$
(d) $9\sin(3x+5)$
›Reveal solutionSolution

Differentiate twice by the chain rule; each derivative brings down a factor of 3, and differentiating sine twice turns it into $-\sin$ .

Let $y=\sin(3x+5)$ .

First derivative: $\dfrac{dy}{dx}=3\cos(3x+5)$ .

…

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Q.If y=ax+by = \sqrt{ax + b}y=ax+b​, prove that yd2ydx2+(dydx)2=0y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0ydx2d2y​+(dxdy​)2=0.

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Q.If $y = \sqrt{ax + b}$ , prove that $y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$ .

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