If \(y = \displaystyle\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\), then
\(\left(x^2-1\right)^{3/2}\displaystyle\frac{d^2y}{dx^2} =\)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-1\)
Rationalising the denominator, with the numerator and the denominator multiplied by \(\sqrt{x+1}+\sqrt{x-1}\):
\( \displaystyle y = \frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{(x+1)-(x-1)} \)
\( \displaystyle = \frac{2x + 2\sqrt{x^2-1}}{2} \)
\( = x + \sqrt{x^2-1} \)
Differentiating:
\( \displaystyle \frac{dy}{dx} = 1 + \frac{x}{\sqrt{x^2-1}} \)
Differentiating again by the quotient rule:
\( \displaystyle \frac{d^2y}{dx^2} = \frac{\sqrt{x^2-1} - x \cdot \frac{x}{\sqrt{x^2-1}}}{x^2-1} \)
\( \displaystyle = \frac{(x^2-1) - x^2}{\left(x^2-1\right)^{3/2}} \)
\( \displaystyle = \frac{-1}{\left(x^2-1\right)^{3/2}} \)
Hence
\( \displaystyle \left(x^2-1\right)^{3/2}\frac{d^2y}{dx^2} = -1 \)