Board 2023 Class 12 Maths Q24 Solution: Show that (x² − 1)(dy/dx)² = 4y² for y = (x + √(x² − 1))²

The question

If $y = \left(x + \sqrt{x^2 - 1}\right)^2$ ⁠, then show that $(x^2 - 1)\left(\dfrac{dy}{dx}\right)^2 = 4y^2$ ⁠.

Exam answer

$y = \left(x + \sqrt{x^2 - 1}\right)^2$ ⁠. Differentiating w.r.t. $x$ ⁠,

\begin{aligned} \frac{dy}{dx} &= 2\left(x + \sqrt{x^2 - 1}\right) \\ &\quad \times\left(1 + \frac{x}{\sqrt{x^2 - 1}}\right) \end{aligned}

Why?

$y$ is a bracket squared, so use the chain rule with $u = x + \sqrt{x^2 - 1}$ ⁠:

\frac{d}{dx}\left(u^2\right) = 2u\frac{du}{dx}

Inside $u$ ⁠, the root needs the chain rule again:

\begin{aligned} &\frac{d}{dx}\sqrt{x^2 - 1} \\ &= \frac{1}{2\sqrt{x^2 - 1}} \times \frac{d}{dx}(x^2 - 1) \\ &= \frac{1}{2\sqrt{x^2 - 1}} \times 2x \\ &= \frac{2x}{2\sqrt{x^2 - 1}} \\ &= \frac{x}{\sqrt{x^2 - 1}} \end{aligned}

(The outside part is $\sqrt{t}$ ⁠, with derivative $\frac{1}{2\sqrt t}$ ⁠. The inside part is $t = x^2 - 1$ ⁠, with derivative $2x$ ⁠. The 2 cancels.)

So the derivative of the bracket is

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

\begin{aligned} \frac{dy}{dx} &= 2\left(x + \sqrt{x^2 - 1}\right) \\ &\quad \times\frac{\sqrt{x^2 - 1} + x}{\sqrt{x^2 - 1}} \end{aligned}

Why?

Write the second bracket as one fraction over $\sqrt{x^2 - 1}$ ⁠:

\begin{aligned} &1 + \frac{x}{\sqrt{x^2 - 1}} \\ &= \frac{\sqrt{x^2 - 1} + x}{\sqrt{x^2 - 1}} \end{aligned}

The new numerator $\sqrt{x^2 - 1} + x$ is the same as the first bracket $x + \sqrt{x^2 - 1}$ ⁠. Spotting this is the key to the proof.

\begin{aligned} \frac{dy}{dx} &= \frac{2\left(x + \sqrt{x^2 - 1}\right)^2}{\sqrt{x^2 - 1}} \\ &= \frac{2y}{\sqrt{x^2 - 1}} \end{aligned}

Why?

The two equal brackets multiply to a square, and that square is the given $y$ ⁠:

\begin{aligned} &2\left(x + \sqrt{x^2 - 1}\right) \\ &\quad \times \frac{x + \sqrt{x^2 - 1}}{\sqrt{x^2 - 1}} \\ &= \frac{2\left(x + \sqrt{x^2 - 1}\right)^2}{\sqrt{x^2 - 1}} \end{aligned}

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

The target is written with $y$ ⁠, so writing the derivative in terms of $y$ is the right move.

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

Squaring both sides,

Why?

Multiplying by $\sqrt{x^2 - 1}$ cleared the fraction. The target has the square of $\frac{dy}{dx}$ and no root, so now square both sides:

\begin{aligned} &\left(\sqrt{x^2 - 1}\,\frac{dy}{dx}\right)^2 = (2y)^2 \\ &\left(\sqrt{x^2 - 1}\right)^2\left(\frac{dy}{dx}\right)^2 \\ &\quad = (2y)^2 \end{aligned}

\begin{aligned} \left(\sqrt{x^2 - 1}\right)^2 &= x^2 - 1 \\ (2y)^2 &= 4y^2 \end{aligned}

This is the required result, so the answer ends with "Hence proved".

$(x^2 - 1)\left(\dfrac{dy}{dx}\right)^2 = 4y^2$ ⁠. Hence proved.

Where this comes from

Chapter 5, Continuity and Differentiability: Chain rule: show a derivative relation.

Question type: The chain rule

How to spot it: click to see

Type 4. The chain rule: a function inside a function

Spot it: something other than a plain $x$ inside $\sin$ ⁠, $\cos$ ⁠, $\log$ ⁠, $e$ or a power, like $\sin(x^2 + 5)$ or $e^{\cos x}$ ⁠.

Here: $y$ is the square of $x + \sqrt{x^2 - 1}$ ⁠, and that bracket holds a root of $x^2 - 1$ ⁠: differentiate the square, times the derivative of the bracket, $1 + \frac{x}{\sqrt{x^2 - 1}}$ ⁠. This simplifies to $\frac{2y}{\sqrt{x^2 - 1}}$ ⁠; then square.

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Q24: Show that (x2−1)(dydx)2=4y2(x^2 - 1)\left(\frac{dy}{dx}\right)^2 = 4y^2(x2−1)(dxdy​)2=4y2 for y=(x+x2−1)2y = \left(x + \sqrt{x^2 - 1}\right)^2y=(x+x2−1​)2

Q24: Show that $(x^2 - 1)\left(\frac{dy}{dx}\right)^2 = 4y^2$ for $y = \left(x + \sqrt{x^2 - 1}\right)^2$