If \(x^2-y^2 = 1\), then which of the following is correct?
(A) \((x^2 - 1)\left(\displaystyle\frac{dy}{dx}\right)^2=x^2\)
(B) \((x^2-1)\left(\displaystyle\frac{d^2y}{dx^2}\right)^2=x^2\)
(C) \((x^2-1)^3\left(\displaystyle\frac{d^2y}{dx^2}\right)^2=x^2\)
(D) \((x^2-1)^3\left(\displaystyle\frac{d^2y}{dx^2}\right)^2=1\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (D) only
Differentiating \(x^2 - y^2 = 1\) implicitly with respect to x:
\( \displaystyle 2x - 2y\frac{dy}{dx} = 0 \)
\( \displaystyle \frac{dy}{dx} = \frac{x}{y} \)
Differentiating again by the quotient rule:
\( \displaystyle \frac{d^2y}{dx^2} = \frac{y - x\,\displaystyle\frac{dy}{dx}}{y^2} \)
\( \displaystyle \frac{d^2y}{dx^2} = \frac{y - \displaystyle\frac{x^2}{y}}{y^2} \)
\( \displaystyle \frac{d^2y}{dx^2} = \frac{y^2 - x^2}{y^3} \)
Since \(y^2 - x^2 = -1\):
\( \displaystyle \frac{d^2y}{dx^2} = -\frac{1}{y^3} \)
Also \(x^2 - 1 = y^2\), which is used in every statement.
So \(\frac{dy}{dx} = \frac{x}{y}\), \(\frac{d^2y}{dx^2} = -\frac{1}{y^3}\) and \(x^2 - 1 = y^2\).
Checking the statements
(A) \((x^2 - 1)\left(\frac{dy}{dx}\right)^2 = x^2\) — Correct. Substituting:
\( \displaystyle (x^2 - 1)\left(\frac{dy}{dx}\right)^2 = y^2\cdot\frac{x^2}{y^2} = x^2 \)
(B) \((x^2 - 1)\left(\frac{d^2y}{dx^2}\right)^2 = x^2\) — Incorrect. Substituting:
\( \displaystyle (x^2 - 1)\left(\frac{d^2y}{dx^2}\right)^2 = y^2\cdot\frac{1}{y^6} = \frac{1}{(x^2 - 1)^2} \)
(C) \((x^2 - 1)^3\left(\frac{d^2y}{dx^2}\right)^2 = x^2\) — Incorrect. The left side is \(y^6\cdot\frac{1}{y^6} = 1\), not \(x^2\).
(D) \((x^2 - 1)^3\left(\frac{d^2y}{dx^2}\right)^2 = 1\) — Correct. Substituting:
\( \displaystyle (x^2 - 1)^3\left(\frac{d^2y}{dx^2}\right)^2 = y^6\cdot\frac{1}{y^6} = 1 \)