Найти y' (корень 3) если y=x/1-корень x^2+1

Y' = (1-√ (x²+1) - x * (-2x/2√ (x²+1)) / (1-√ (x²+1)) ² = (1-√ (x²+1) - x²/√ (x²+1)) / (1-√ (x²+1)) ²=

= (√ (x²+1) - x²-1-x²) / √ (x²+1) (1-√ (x²+1)) = (√ (x²+1) - 2x²-1) / √ (x²+1) (1-√ (x²+1)) ²

y' (√3) = (2-6-1) / 2*1=-5/2=-2,5