If $x^2 - y^2 = 1$, then which of the following is correct?

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

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

(C) $(x^2 - 1)^3\left(\frac{d^2y}{dx^2}\right)^2 = x^2$

(D) $(x^2 - 1)^3\left(\frac{d^2y}{dx^2}\right)^2 = 1$

Choose the correct answer from the options given below:

Given the equation $x^2 - y^2 = 1$, we need to determine which statement about derivatives is correct.

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Differentiating both sides with respect to $x$:

$2x - 2y\frac{dy}{dx} = 0$

$2x = 2y\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{x}{y}$

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Differentiating $\frac{dy}{dx} = \frac{x}{y}$ using the quotient rule:

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

Substituting $\frac{dy}{dx} = \frac{x}{y}$:

$\frac{d^2y}{dx^2} = \frac{y - x \cdot \frac{x}{y}}{y^2}$

$\frac{d^2y}{dx^2} = \frac{y - \frac{x^2}{y}}{y^2}$

$\frac{d^2y}{dx^2} = \frac{y^2 - x^2}{y^3}$

Since $x^2 - y^2 = 1$, we have $y^2 - x^2 = -1$:

$\frac{d^2y}{dx^2} = \frac{-1}{y^3}$

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From $x^2 - y^2 = 1$:

$y^2 = x^2 - 1$

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Checking statement (A): $(x^2 - 1)\left(\frac{dy}{dx}\right)^2 = x^2$

$(x^2 - 1)\left(\frac{x}{y}\right)^2 = (x^2 - 1) \cdot \frac{x^2}{y^2}$

Substituting $y^2 = x^2 - 1$:

$= (x^2 - 1) \cdot \frac{x^2}{x^2 - 1}$

$= x^2$

Statement (A) is correct.

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Checking statement (B): $(x^2 - 1)\left(\frac{d^2y}{dx^2}\right)^2 = x^2$

$(x^2 - 1)\left(\frac{-1}{y^3}\right)^2 = (x^2 - 1) \cdot \frac{1}{y^6}$

$= (x^2 - 1) \cdot \frac{1}{(y^2)^3}$

$= (x^2 - 1) \cdot \frac{1}{(x^2 - 1)^3}$

$= \frac{1}{(x^2 - 1)^2} \neq x^2$

Statement (B) is incorrect.

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Checking statement (C): $(x^2 - 1)^3\left(\frac{d^2y}{dx^2}\right)^2 = x^2$

$(x^2 - 1)^3 \cdot \frac{1}{y^6} = (x^2 - 1)^3 \cdot \frac{1}{(x^2 - 1)^3}$

$= 1 \neq x^2$

Statement (C) is incorrect.

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Checking statement (D): $(x^2 - 1)^3\left(\frac{d^2y}{dx^2}\right)^2 = 1$

$(x^2 - 1)^3 \cdot \frac{1}{(x^2 - 1)^3} = 1$

Statement (D) is correct.

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Therefore, statements (A) and (D) are correct.