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CBSE Board Examination 2025 · Section D · 5 marks

Q32: Prove dydx=1−y21−x2\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}⁠, or d2ydx2\frac{d^2y}{dx^2} at θ=π4\theta = \frac{\pi}{4} for a parametric curve

Board 2025Ch 5 · Continuity and Differentiability
The question

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

OR

(b) If x=a(cos⁡θ+log⁡tan⁡θ2)x = a\left(\cos\theta + \log\tan\dfrac{\theta}{2}\right) and y=sin⁡θy = \sin\theta⁠, then find d2ydx2\dfrac{d^2y}{dx^2} at θ=π4\theta = \dfrac{\pi}{4}⁠.

Exam answer

(a)

Put x=sin⁡Ax = \sin A and y=sin⁡By = \sin B⁠, so A=sin⁡−1xA = \sin^{-1}x and B=sin⁡−1yB = \sin^{-1}y⁠. Then 1−x2=cos⁡A\sqrt{1 - x^2} = \cos A and 1−y2=cos⁡B\sqrt{1 - y^2} = \cos B⁠.

Why?

The roots 1−x2\sqrt{1 - x^2} and 1−y2\sqrt{1 - y^2} suggest a sine substitution, because 1−sin⁡2A=cos⁡2A1 - \sin^2A = \cos^2A⁠.

With A=sin⁡−1xA = \sin^{-1}x⁠, AA lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, where cos⁡A≥0\cos A \ge 0⁠. So cos⁡2A=cos⁡A\sqrt{\cos^2A} = \cos A⁠, with no sign trouble. The same holds for BB⁠.

cos⁡A+cos⁡B=a(sin⁡A−sin⁡B)\cos A + \cos B = a(\sin A - \sin B) 2cos⁡A+B2cos⁡A−B2=2acos⁡A+B2sin⁡A−B2\begin{aligned} &2\cos\frac{A + B}{2}\cos\frac{A - B}{2} \\ &= 2a\cos\frac{A + B}{2}\sin\frac{A - B}{2} \end{aligned}
Why?

Use the sum-to-product formulas on each side:

cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\begin{aligned} &\cos C + \cos D \\ &= 2\cos\frac{C + D}{2}\cos\frac{C - D}{2} \end{aligned} sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2\begin{aligned} &\sin C - \sin D \\ &= 2\cos\frac{C + D}{2}\sin\frac{C - D}{2} \end{aligned}

Both sides now share the factor 2cos⁡A+B22\cos\frac{A + B}{2}⁠.

Dividing both sides by 2cos⁡A+B2sin⁡A−B22\cos\frac{A + B}{2}\sin\frac{A - B}{2} (non-zero for ∣x∣<1|x| < 1⁠),

Why?

Cancelling 2cos⁡A+B22\cos\frac{A + B}{2} and dividing by sin⁡A−B2\sin\frac{A - B}{2} leaves cos⁡sin⁡=cot⁡\frac{\cos}{\sin} = \cot on the left.

2cos⁡A+B2cos⁡A−B2=2acos⁡A+B2sin⁡A−B2\begin{aligned} &2\cos\tfrac{A + B}{2}\cos\tfrac{A - B}{2} \\ &= 2a\cos\tfrac{A + B}{2}\sin\tfrac{A - B}{2} \end{aligned}

Divide both sides by 2cos⁡A+B22\cos\frac{A + B}{2}⁠:

cos⁡A−B2=asin⁡A−B2\begin{aligned} \cos\tfrac{A - B}{2} &= a\sin\tfrac{A - B}{2} \end{aligned}

Divide both sides by sin⁡A−B2\sin\frac{A - B}{2}⁠:

cos⁡A−B2sin⁡A−B2=acot⁡A−B2=a\begin{aligned} \frac{\cos\frac{A - B}{2}}{\sin\frac{A - B}{2}} &= a \\ \cot\tfrac{A - B}{2} &= a \end{aligned}

Then take cot⁡−1\cot^{-1} of both sides and multiply by 2: A−B=2cot⁡−1aA - B = 2\cot^{-1}a⁠.

The divisor is not zero. cos⁡A+B2=0\cos\frac{A + B}{2} = 0 would need A=B=±π2A = B = \pm\frac{\pi}{2}⁠, so ∣x∣=1|x| = 1⁠. sin⁡A−B2=0\sin\frac{A - B}{2} = 0 would need x=yx = y⁠, and then the given relation says 21−x2=02\sqrt{1 - x^2} = 0⁠, so again ∣x∣=1|x| = 1⁠.

cot⁡A−B2=a⇒A−B=2cot⁡−1a\begin{aligned} \cot\frac{A - B}{2} &= a \\ \Rightarrow A - B &= 2\cot^{-1}a \end{aligned} ⇒sin⁡−1x−sin⁡−1y=2cot⁡−1a\Rightarrow \sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a
Why?

The key fact is that A−BA - B is a constant, since aa is a constant. When x<yx < y⁠, A−B2\frac{A - B}{2} is negative and equals cot⁡−1a−π\cot^{-1}a - \pi⁠. It is still a constant, so the derivative below is the same.

Differentiating both sides w.r.t. xx⁠,

11−x2−11−y2dydx=0\frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}}\frac{dy}{dx} = 0
Why?

The standard derivative is

ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}

For sin⁡−1y\sin^{-1}y use the chain rule: yy depends on xx⁠, so multiply by dydx\frac{dy}{dx}⁠. The right side 2cot⁡−1a2\cot^{-1}a is a constant, so its derivative is 00⁠.

11−y2dydx=11−x2\frac{1}{\sqrt{1 - y^2}}\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}
Why?

Move the dydx\frac{dy}{dx} term to the other side, then multiply both sides by 1−y2\sqrt{1 - y^2}⁠. Two square roots divide as one (for m≥0m \ge 0⁠, n>0n > 0⁠):

11−y2dydx=11−x2dydx=1−y2⋅11−x2=1−y21−x2\begin{aligned} \frac{1}{\sqrt{1 - y^2}}\frac{dy}{dx} &= \frac{1}{\sqrt{1 - x^2}} \\ \frac{dy}{dx} &= \sqrt{1 - y^2}\cdot\frac{1}{\sqrt{1 - x^2}} \\ &= \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \end{aligned} mn=mn\frac{\sqrt m}{\sqrt n} = \sqrt{\frac{m}{n}}

∴ dydx=1−y21−x2=1−y21−x2\therefore\ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} = \sqrt{\frac{1 - y^2}{1 - x^2}}⁠. Hence proved.

(b)

dxdθ=a(−sin⁡θ+sec⁡2θ2⋅12tan⁡θ2)=a(−sin⁡θ+12sin⁡θ2cos⁡θ2)=a(−sin⁡θ+1sin⁡θ)=a⋅1−sin⁡2θsin⁡θ=acos⁡2θsin⁡θ\begin{aligned} \frac{dx}{d\theta} &= a\left(-\sin\theta + \frac{\sec^2\frac{\theta}{2}\cdot\frac12}{\tan\frac{\theta}{2}}\right) \\ &= a\left(-\sin\theta + \frac{1}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right) \\ &= a\left(-\sin\theta + \frac{1}{\sin\theta}\right) \\ &= a\cdot\frac{1 - \sin^2\theta}{\sin\theta} = \frac{a\cos^2\theta}{\sin\theta} \end{aligned}
Why?

xx and yy are both given in terms of θ\theta (parametric form), so differentiate each with respect to θ\theta⁠.

For log⁡tan⁡θ2\log\tan\frac{\theta}{2}⁠, use the chain rule twice:

ddθlog⁡u=1ududθddθtan⁡θ2=12sec⁡2θ2\begin{aligned} \frac{d}{d\theta}\log u &= \frac{1}{u}\frac{du}{d\theta} \\ \frac{d}{d\theta}\tan\frac{\theta}{2} &= \frac12\sec^2\frac{\theta}{2} \end{aligned}

To simplify, write sec⁡2\sec^2 and tan⁡\tan in sines and cosines:

sec⁡2θ2tan⁡θ2=1sin⁡θ2cos⁡θ2\frac{\sec^2\frac{\theta}{2}}{\tan\frac{\theta}{2}} = \frac{1}{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}

In detail, with u=tan⁡θ2u = \tan\frac{\theta}{2} and short names s=sin⁡θ2s = \sin\frac{\theta}{2}⁠, c=cos⁡θ2c = \cos\frac{\theta}{2}⁠:

ddθlog⁡tan⁡θ2=1tan⁡θ2⋅12sec⁡2θ2\begin{aligned} &\frac{d}{d\theta}\log\tan\tfrac{\theta}{2} \\ &= \frac{1}{\tan\frac{\theta}{2}}\cdot\frac12\sec^2\tfrac{\theta}{2} \end{aligned} sec⁡2θ2tan⁡θ2=1c2÷sc=1c2⋅cs=1sc12⋅1sc=12sc=1sin⁡θ\begin{aligned} \frac{\sec^2\frac{\theta}{2}}{\tan\frac{\theta}{2}} &= \frac{1}{c^2} \div \frac{s}{c} \\ &= \frac{1}{c^2}\cdot\frac{c}{s} = \frac{1}{sc} \\[4pt] \frac12\cdot\frac{1}{sc} &= \frac{1}{2sc} = \frac{1}{\sin\theta} \end{aligned}

Then 2sin⁡θ2cos⁡θ2=sin⁡θ2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = \sin\theta⁠, and 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta⁠.

Put the two terms over the common denominator sin⁡θ\sin\theta⁠:

−sin⁡θ+1sin⁡θ=−sin⁡2θsin⁡θ+1sin⁡θ=1−sin⁡2θsin⁡θ=cos⁡2θsin⁡θ\begin{aligned} &-\sin\theta + \frac{1}{\sin\theta} \\ &= \frac{-\sin^2\theta}{\sin\theta} + \frac{1}{\sin\theta} \\ &= \frac{1 - \sin^2\theta}{\sin\theta} \\ &= \frac{\cos^2\theta}{\sin\theta} \end{aligned}
dydθ=cos⁡θ\frac{dy}{d\theta} = \cos\theta dydx=dy/dθdx/dθ=cos⁡θsin⁡θacos⁡2θ=tan⁡θa\begin{aligned} \frac{dy}{dx} &= \frac{dy/d\theta}{dx/d\theta} = \frac{\cos\theta\sin\theta}{a\cos^2\theta} \\ &= \frac{\tan\theta}{a} \end{aligned}
Why?

For a parametric curve, divide the two derivatives:

dydx=dy/dθdx/dθ\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}

Dividing by acos⁡2θsin⁡θ\frac{a\cos^2\theta}{\sin\theta} means multiplying by sin⁡θacos⁡2θ\frac{\sin\theta}{a\cos^2\theta}⁠. One cos⁡θ\cos\theta cancels, leaving sin⁡θacos⁡θ\frac{\sin\theta}{a\cos\theta}⁠.

dydx=cos⁡θ÷acos⁡2θsin⁡θ=cos⁡θ⋅sin⁡θacos⁡2θ=sin⁡θacos⁡θ=1a⋅sin⁡θcos⁡θ=tan⁡θa\begin{aligned} \frac{dy}{dx} &= \cos\theta \div \frac{a\cos^2\theta}{\sin\theta} \\ &= \cos\theta\cdot\frac{\sin\theta}{a\cos^2\theta} \\ &= \frac{\sin\theta}{a\cos\theta} \\ &= \frac{1}{a}\cdot\frac{\sin\theta}{\cos\theta} = \frac{\tan\theta}{a} \end{aligned}

Differentiating w.r.t. xx⁠,

d2ydx2=sec⁡2θa⋅dθdx=sec⁡2θa⋅sin⁡θacos⁡2θ=sec⁡3θtan⁡θa2\begin{aligned} \frac{d^2y}{dx^2} &= \frac{\sec^2\theta}{a}\cdot\frac{d\theta}{dx} \\ &= \frac{\sec^2\theta}{a}\cdot\frac{\sin\theta}{a\cos^2\theta} \\ &= \frac{\sec^3\theta\tan\theta}{a^2} \end{aligned}
Why?

dydx=tan⁡θa\frac{dy}{dx} = \frac{\tan\theta}{a} is a function of θ\theta⁠. To differentiate it with respect to xx⁠, differentiate with respect to θ\theta and multiply by dθdx\frac{d\theta}{dx} (chain rule). Here

dθdx=1dx/dθ=sin⁡θacos⁡2θ\frac{d\theta}{dx} = \frac{1}{dx/d\theta} = \frac{\sin\theta}{a\cos^2\theta}

Then sin⁡θcos⁡2θ=sec⁡θtan⁡θ\frac{\sin\theta}{\cos^2\theta} = \sec\theta\tan\theta⁠, which gives sec⁡3θtan⁡θ\sec^3\theta\tan\theta⁠.

ddθ(tan⁡θa)=sec⁡2θasin⁡θcos⁡2θ=1cos⁡θ⋅sin⁡θcos⁡θ=sec⁡θtan⁡θ\begin{aligned} \frac{d}{d\theta}\left(\frac{\tan\theta}{a}\right) &= \frac{\sec^2\theta}{a} \\[4pt] \frac{\sin\theta}{\cos^2\theta} &= \frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta} \\ &= \sec\theta\tan\theta \end{aligned} sec⁡2θa⋅sec⁡θtan⁡θa=sec⁡3θtan⁡θa2\begin{aligned} &\frac{\sec^2\theta}{a}\cdot\frac{\sec\theta\tan\theta}{a} \\ &= \frac{\sec^3\theta\tan\theta}{a^2} \end{aligned}

Watch out: d2ydx2\frac{d^2y}{dx^2} is not d2ydθ2\frac{d^2y}{d\theta^2} divided by d2xdθ2\frac{d^2x}{d\theta^2}⁠.

At θ=π4\theta = \frac{\pi}{4}⁠, sec⁡π4=2\sec\frac{\pi}{4} = \sqrt2 and tan⁡π4=1\tan\frac{\pi}{4} = 1⁠, so

d2ydx2=(2)3⋅1a2=22a2\frac{d^2y}{dx^2} = \frac{(\sqrt2)^3\cdot1}{a^2} = \frac{2\sqrt2}{a^2}
Why?

cos⁡π4=12\cos\frac{\pi}{4} = \frac{1}{\sqrt2}⁠, so sec⁡π4=2\sec\frac{\pi}{4} = \sqrt2⁠. Also (2)3=2⋅2⋅2=22(\sqrt2)^3 = \sqrt2\cdot\sqrt2\cdot\sqrt2 = 2\sqrt2⁠.

∴ d2ydx2=22a2\therefore\ \dfrac{d^2y}{dx^2} = \dfrac{2\sqrt2}{a^2} at θ=π4\theta = \dfrac{\pi}{4}⁠.

Where this comes from

Chapter 5, Continuity and Differentiability: Implicit, parametric and second derivatives.

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

How to spot it: click to see

(a) Type 5. Implicit: xx and yy mixed in one equation

Spot it: an equation such as x2+xy+y2=100x^2 + xy + y^2 = 100 that you cannot easily solve for yy⁠.

Here: xx and yy are mixed under two roots. Put x=sin⁡Ax = \sin A⁠, y=sin⁡By = \sin B to reach sin⁡−1x−sin⁡−1y=2cot⁡−1a\sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a⁠, then differentiate both sides: the yy term brings dydx\frac{dy}{dx}⁠.

(b) Type 8. Parametric form: xx and yy given through tt or θ\theta

Spot it: two equations, one for xx and one for yy⁠, both in the same third letter.

Here: First, xx and yy are both given in θ\theta⁠: dxdθ=acos⁡2θsin⁡θ\frac{dx}{d\theta} = \frac{a\cos^2\theta}{\sin\theta} and dydθ=cos⁡θ\frac{dy}{d\theta} = \cos\theta⁠, so dydx=tan⁡θa\frac{dy}{dx} = \frac{\tan\theta}{a}⁠.

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

Spot it: d2ydx2\frac{d^2y}{dx^2} or y′′y'' in the question, often "prove that d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0".

Here: Then differentiate tan⁡θa\frac{\tan\theta}{a} again with respect to xx⁠: that gives sec⁡2θa⋅dθdx\frac{\sec^2\theta}{a}\cdot\frac{d\theta}{dx}⁠. Put θ=π4\theta = \frac\pi4 at the end.

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