Question

y_(2) :- d^2y/dx^2
y_(1) :- dy/dx

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Answer 1

Step-by-step explanation:

We have,

$2y = \sqrt{x + 1} + \sqrt{x - 1}$

Differentiating both sides,

$2 \: y_{1} = \frac{1}{2 \sqrt{x + 1} } + \frac{1}{2 \sqrt{x - 1} }$

$= > 2 \: y_{1} = \frac{ \sqrt{x + 1} + \sqrt{x - 1} }{2 \sqrt{ {x}^{2} - 1 } }$

$= > 2 \: y_{1} = \frac{2y}{2 \sqrt{ {x}^{2} - 1 } }$

$= > 2 \: y_{1} = \frac{y}{ \sqrt{ {x}^{2} - 1} }$

Differentiating both sides,