CBSE Sample Paper 2026-27 · Section C · 3 marks

Q28: $\frac{d^2y}{dx^2}$ for $x = \sin^3t$ ⁠, $y = \cos^3t$ ⁠; or $\frac{dy}{dx}$ for $y = x^{\log x} + (\log x)^x$

Sample 2026-27Ch 5 · Continuity and Differentiability

The question

(a) If $x = \sin^3t$ ⁠, $y = \cos^3t$ ⁠, Evaluate $\dfrac{d^2y}{dx^2}$ at $\left(\dfrac{1}{2\sqrt2}, \dfrac{1}{2\sqrt2}\right)$ ⁠.

(b) If $y = x^{\log x} + (\log x)^x$ ; $x > 1$ ⁠, then evaluate $\dfrac{dy}{dx}$

Exam answer

(a)

\begin{aligned} \frac{dx}{dt} &= 3\sin^2t\cos t \\ \frac{dy}{dt} &= -3\cos^2t\sin t \end{aligned}

Why?

Chain rule on a power: bring the power down, then multiply by the derivative of the inside.

\begin{aligned} \frac{d}{dt}\sin^3t &= 3\sin^2t\cdot\cos t \\ \frac{d}{dt}\cos^3t &= 3\cos^2t\cdot(-\sin t) \end{aligned}

Watch the minus sign: $\frac{d}{dt}\cos t = -\sin t$ ⁠.

\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} = \frac{-3\cos^2t\sin t}{3\sin^2t\cos t} \\ &= -\cot t \end{aligned}

Why?

For a curve given by a parameter $t$ ⁠, divide the two $t$ -derivatives.

Cancel $3\sin t\cos t$ from the top and bottom. That leaves $-\frac{\cos t}{\sin t} = -\cot t$ ⁠.

\begin{aligned} &\frac{-3\cos^2t\sin t}{3\sin^2t\cos t} \\ &= \frac{-(3\sin t\cos t)\cos t}{(3\sin t\cos t)\sin t} \\ &= -\frac{\cos t}{\sin t} = -\cot t \end{aligned}

\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx} \\ &= \operatorname{cosec}^2t\cdot\frac{1}{3\sin^2t\cos t} \\ &= \frac13\sec t\operatorname{cosec}^4t \end{aligned}

Why?

$\frac{dy}{dx}$ is in terms of $t$ ⁠, so differentiate it with respect to $t$ and multiply by $\frac{dt}{dx}$ (chain rule).

\begin{aligned} \frac{d}{dt}(-\cot t) &= -(-\operatorname{cosec}^2t) \\ &= \operatorname{cosec}^2t \\ \frac{dt}{dx} &= \frac{1}{dx/dt} \\ &= \frac{1}{3\sin^2t\cos t} \end{aligned}

Then write $\operatorname{cosec}^2t = \frac{1}{\sin^2t}$ and multiply:

\begin{aligned} &\frac{1}{\sin^2t}\cdot\frac{1}{3\sin^2t\cos t} \\ &= \frac13\cdot\frac{1}{\cos t}\cdot\frac{1}{\sin^4t} \end{aligned}

and $\frac{1}{\cos t} = \sec t$ ⁠, $\frac{1}{\sin t} = \operatorname{cosec}t$ ⁠.

Watch out: $\frac{d^2y}{dx^2}$ is not $\frac{d^2y}{dt^2}$ divided by $\frac{d^2x}{dt^2}$ ⁠.

At the point $\left(\frac{1}{2\sqrt2}, \frac{1}{2\sqrt2}\right)$ ⁠, $\sin^3t = \cos^3t = \frac{1}{2\sqrt2}$ ⁠, so $\sin t = \cos t = \frac{1}{\sqrt2}$ and $t = \frac{\pi}{4}$ ⁠.

Why?

$2\sqrt2 = (\sqrt2)^3$ ⁠, so

(\sqrt2)^3 = (\sqrt2)^2 \times \sqrt2 = 2\sqrt2

\frac{1}{2\sqrt2} = \left(\frac{1}{\sqrt2}\right)^3

Taking cube roots gives $\sin t = \cos t = \frac{1}{\sqrt2}$ ⁠. Both are positive and equal at $t = \frac{\pi}{4}$ ⁠.

\frac{d^2y}{dx^2}\bigg|_{t = \frac{\pi}{4}} = \frac13\cdot\sqrt2\cdot\left(\sqrt2\right)^4

= \frac13\cdot\sqrt2\cdot 4 = \frac{4\sqrt2}{3}

Why?

At $t = \frac{\pi}{4}$ ⁠: $\sec t = \frac{1}{\cos t} = \sqrt2$ and $\operatorname{cosec}t = \sqrt2$ ⁠.

\begin{aligned} \sec\frac{\pi}{4} &= \frac{1}{1/\sqrt2} = \sqrt2 \\ \operatorname{cosec}\frac{\pi}{4} &= \frac{1}{1/\sqrt2} = \sqrt2 \\ \operatorname{cosec}^4\frac{\pi}{4} &= (\sqrt2)^4 \end{aligned}

$\left(\sqrt2\right)^4 = \left(\left(\sqrt2\right)^2\right)^2 = 2^2 = 4$ ⁠.

\begin{aligned} \frac13\cdot\sqrt2\cdot 4 &= \frac{1 \times 4\sqrt2}{3} \\ &= \frac{4\sqrt2}{3} \end{aligned}

$\therefore\ \frac{d^2y}{dx^2} = \frac{4\sqrt2}{3}$ at the given point.

(b)

Let $u = x^{\log x}$ and $v = (\log x)^x$ ⁠, so $y = u + v$ ⁠.

Why?

Both terms have the variable in the power, so each needs logarithmic differentiation.

Watch out: do not take log of the whole of $y$ ⁠. There is no rule for the log of a sum. Split $y$ into $u + v$ first and treat each part separately.

\begin{aligned} \log u &= (\log x)^2 \\ \Rightarrow \frac1u\frac{du}{dx} &= 2\log x\cdot\frac1x \\ \Rightarrow \frac{du}{dx} &= x^{\log x}\cdot\frac{2\log x}{x} \end{aligned}

Why?

Take log of both sides and use $\log a^b = b\log a$ ⁠:

\log x^{\log x} = \log x\cdot\log x

Differentiate. The left side gives $\frac1u\frac{du}{dx}$ ⁠. On the right, the chain rule gives $2\log x\cdot\frac1x$ ⁠. Then multiply by $u = x^{\log x}$ ⁠.

\begin{aligned} \frac{d}{dx}(\log x)^2 &= 2(\log x)^1\cdot\frac{d}{dx}\log x \\ &= 2\log x\cdot\frac1x \\[4pt] \frac{du}{dx} &= u\cdot\frac{2\log x}{x} \\ &= x^{\log x}\cdot\frac{2\log x}{x} \end{aligned}

\begin{aligned} \log v &= x\log(\log x) \\ \Rightarrow \frac1v\frac{dv}{dx} &= 1\cdot\log(\log x) \\ &\quad + x\cdot\frac{1}{\log x}\cdot\frac1x \\ \Rightarrow \frac1v\frac{dv}{dx} &= \log(\log x) + \frac{1}{\log x} \\ \Rightarrow \frac{dv}{dx} &= (\log x)^x\bigg(\log(\log x) \\ &\qquad + \frac{1}{\log x}\bigg) \end{aligned}

Why?

The right side $x\cdot\log(\log x)$ is a product, so use the product rule.

\begin{aligned} \text{first} &= x,\quad \text{its derivative} = 1 \\ \text{second} &= \log(\log x) \end{aligned}

Derivative = (derivative of first) $\times$ second $+$ first $\times$ (derivative of second).

The derivative of $\log(\log x)$ needs the chain rule:

\frac{d}{dx}\log(\log x) = \frac{1}{\log x}\cdot\frac1x

Then $x\cdot\frac{1}{x\log x} = \frac{1}{\log x}$ ⁠. Finally multiply by $v = (\log x)^x$ ⁠.

\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}

Why?

$y = u + v$ ⁠, so the derivative of the sum is the sum of the derivatives. Put in the two results found above:

\begin{aligned} \frac{du}{dx} &= \frac{2\log x}{x}\,x^{\log x} \\ \frac{dv}{dx} &= (\log x)^x\bigg(\log(\log x) \\ &\qquad + \frac{1}{\log x}\bigg) \end{aligned}

$\therefore\ \dfrac{dy}{dx} = \dfrac{2\log x}{x}\,x^{\log x}$ $+ (\log x)^x\Big(\log(\log x)$ $+ \dfrac{1}{\log x}\Big)$

Where this comes from

Chapter 5, Continuity and Differentiability: Parametric second derivative; log differentiation.

Question type: (a) Parametric form + Second derivative (combined); (b) Logarithmic differentiation

How to spot it: click to see

(a) Type 8. Parametric form: $x$ and $y$ given through $t$ or $\theta$

Spot it: two equations, one for $x$ and one for $y$ ⁠, both in the same third letter.

Here: First, $x = \sin^3t$ and $y = \cos^3t$ share $t$ ⁠: divide to get $\frac{dy}{dx} = -\cot t$ ⁠.

(a) Type 9. Second derivative: differentiate twice, then prove

Spot it: $\frac{d^2y}{dx^2}$ or $y''$ in the question, often "prove that $\frac{d^2y}{dx^2} + y = 0$ ".

Here: Then differentiate $-\cot t$ again, times $\frac{dt}{dx} = \frac{1}{3\sin^2t\cos t}$ ⁠. The point has $\sin^3t = \cos^3t$ ⁠, so $t = \frac\pi4$ ⁠.

(b) Type 7. Logarithmic differentiation: $x$ in the power

Spot it: a power whose exponent contains $x$ ⁠, like $x^x$ ⁠, $x^{\sin x}$ or $(\log x)^{\cos x}$ ⁠, or a long product.

Here: Two powers with $x$ up top, $x^{\log x}$ and $(\log x)^x$ ⁠: name them $u$ and $v$ ⁠, take log of each, $(\log x)^2$ and $x\log(\log x)$ ⁠, and differentiate.

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Continuity and Differentiability in past papers: question 17 of 35 · see them all

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Q28: $\frac{d^2y}{dx^2}$ for $x = \sin^3t$ ⁠, $y = \cos^3t$ ⁠; or $\frac{dy}{dx}$ for $y = x^{\log x} + (\log x)^x$