CBSE Sample Paper 2022-23 · Section A · 1 mark

Q16: $(1 - x^2)y_2$ when $y = \sin^{-1}x$

Sample 2022-23Ch 5 · Continuity and Differentiability

The question

If $y = \sin^{-1}x$ ⁠, then $(1 - x^2)y_2$ is equal to

(a) $xy_1$

(b) $xy$

(d) $x^2$

Exam answer

Differentiating $y = \sin^{-1}x$ w.r.t. x:

y_1 = \frac{1}{\sqrt{1 - x^2}} \Rightarrow \sqrt{1 - x^2}\,y_1 = 1

Why?

The standard derivative is

\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}

Clearing the fraction before differentiating again is the useful trick: the next step is then a simple product rule, not a quotient.

Differentiating again w.r.t. x:

\sqrt{1 - x^2}\,y_2 + \frac{-2x\,y_1}{2\sqrt{1 - x^2}} = 0

Why?

Product rule on $\sqrt{1 - x^2}\cdot y_1$ ⁠: the first factor times $y_2$ ⁠, plus $y_1$ times the derivative of the first factor. By the chain rule,

\frac{d}{dx}\sqrt{1 - x^2} = \frac{-2x}{2\sqrt{1 - x^2}}

The right side is the constant 1, whose derivative is 0.

Multiplying by $\sqrt{1 - x^2}$ ⁠: $(1 - x^2)y_2 - xy_1 = 0$ ⁠.

Why?

In the first term, $\sqrt{1 - x^2}\cdot\sqrt{1 - x^2} = 1 - x^2$ ⁠. In the second, the square roots cancel and $\frac{-2x}{2} = -x$ ⁠.

\begin{aligned} &\sqrt{1 - x^2} \cdot \sqrt{1 - x^2}\,y_2 \\ &= (1 - x^2)y_2 \\[4pt] &\sqrt{1 - x^2} \cdot \frac{-2x\,y_1}{2\sqrt{1 - x^2}} \\ &= \frac{-2x\,y_1}{2} = -xy_1 \\[4pt] &\sqrt{1 - x^2} \cdot 0 = 0 \end{aligned}

Then move $-xy_1$ to the right side: $(1 - x^2)y_2 = xy_1$ ⁠.

$\therefore\ (1 - x^2)y_2 = xy_1$

Why?

Watch out: the answer has $y_1$ ⁠, the first derivative, not $y$ itself. Option (b), $xy$ ⁠, mixes them up.

Answer: (a) $xy_1$

Where this comes from

Chapter 5, Continuity and Differentiability: Second order derivative of sin⁻¹x.

Question type: Second derivative

How to spot it: click to see

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: $y_2$ means differentiate twice. First clear the root, $\sqrt{1 - x^2}\,y_1 = 1$ ⁠, then differentiate again with the product rule and multiply by $\sqrt{1 - x^2}$ ⁠: $(1 - x^2)y_2 = xy_1$ ⁠.

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Continuity and Differentiability in past papers: question 31 of 35 · see them all

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Q16: $(1 - x^2)y_2$ when $y = \sin^{-1}x$