Question

y=√1+x^2/1-x^2 then dy/dx is​

Answer 1

Step-by-step explanation:

$y = \sqrt{1} + \frac{ {x}^{2} }{1} - {x}^{2} \: then \: \frac{dy}{dx} \: is$

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

$y = 1 + \cancel{ {x}^{2} } - \cancel{ {x}^{2} }$

$y = 1$

Differential with respect to x,

$\frac{dy}{dx} = \frac{d}{dx} (1)$

$= 0$

I hope it is helpful