Question

Производные y=$frac{x^3+3}{x^2-1}$ ;$x_{0}$=2

Answer 1

y' = $frac{3x^{2}(x^{2}-1)-(x^{3}+3)2x}{(x^{2}-1)^{2}} = frac{3x^{4}-3x^{2}-2x^{4}-6x}{(x^{2}-1)^{2}} = frac{x^{4}-3x^{2}-6x}{(x^{2}-1)^{2}}$
y'(x₀) = y'(2) = $frac{2^{4}-3*2^{2}-6*2}{(2^{2}-1)^{2}} = frac{16-12-12}{9} = - frac{8}{9}$