2021 H2 Maths Paper 1 Q7 — Worked Solution

Part (a)

4 marks

y = e^{\sin^{- 1}x}

Differentiating using the chain rule, with $\frac{d}{dx}\left( \sin^{- 1}x \right) = \frac{1}{\sqrt{1 - x^{2}}}$ :

\frac{dy}{dx} = e^{\sin^{- 1}x} \times \frac{1}{\sqrt{1 - x^{2}}}

= \frac{y}{\sqrt{1 - x^{2}}}

Approach

this relates $\frac{dy}{dx}$ to $y$ , but the target equation has $\frac{d^{2}y}{dx^{2}}$ , so we need a second derivative. To avoid differentiating $\frac{1}{\sqrt{1 - x^{2}}}$ directly, we first clear the square root by squaring, then differentiate implicitly.

Rearranging and squaring both sides:

\sqrt{1 - x^{2}}\,\frac{dy}{dx} = y

\Rightarrow \left( 1 - x^{2} \right)\left( \frac{dy}{dx} \right)^{2} = y^{2}

Differentiating both sides implicitly with respect to $x$ (product rule on the left, chain rule on the right):

- 2x\left( \frac{dy}{dx} \right)^{2} + \left( 1 - x^{2} \right) \times 2\frac{dy}{dx}\frac{d^{2}y}{dx^{2}} = 2y\frac{dy}{dx}

Dividing throughout by $2\frac{dy}{dx}$ , which is non-zero since $\frac{dy}{dx} = \frac{y}{\sqrt{1 - x^{2}}}$ and $y = e^{\sin^{- 1}x} > 0$ :

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

\left( 1 - x^{2} \right)\frac{d^{2}y}{dx^{2}} = x\frac{dy}{dx} + y\ \ \left( \text{shown} \right)

To check your answer: at

x = 0

y = e^{\sin^{- 1}0} = 1

and

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

; substituting into the shown identity gives

\frac{d^{2}y}{dx^{2}} = 0(1) + 1 = 1

x = 0

, a finite, sensible value. ✓

Part (b)

5 marks

At $x = 0$ : $y(0) = e^{\sin^{- 1}0} = 1$ .

From $\frac{dy}{dx} = \frac{y}{\sqrt{1 - x^{2}}}$ , at $x = 0$ : $y'(0) = 1$ .

From the result of part (a), $\left( 1 - x^{2} \right)y'' = xy'+ y$ ; at $x = 0$ :

y''(0) = 0 \times y'(0) + y(0) = 1

Differentiating $\left( 1 - x^{2} \right)y'' = xy'+ y$ with respect to $x$ (product rule on both sides):

- 2xy'' + \left( 1 - x^{2} \right)y''' = y'+ xy'' + y'

= 2y'+ xy''

\Rightarrow \left( 1 - x^{2} \right)y''' = 2y'+ 3xy''

At $x = 0$ :

y'''(0) = 2y'(0) + 0 = 2

Using the Maclaurin series $y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^{2} + \frac{y'''(0)}{3!}x^{3} + \cdots$ :

y = 1 + x + \frac{1}{2}x^{2} + \frac{2}{6}x^{3} + \cdots

= 1 + x + \frac{1}{2}x^{2} + \frac{1}{3}x^{3} + \cdots

To check your answer: at

x = 0.1

e^{\sin^{- 1}(0.1)} \approx 1.10536

, and

1 + 0.1 + \frac{1}{2}(0.01) + \frac{1}{3}(0.001) \approx 1.10533

, a close match. ✓

$\therefore$ $e^{\sin^{- 1}x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \cdots$

2021 H2 Maths Paper 1, Question 7

Part (a)

Part (b)

This is how we mark, and how we teach.