Question

Вычислить y' в точке х1
у= (2x-1) / (4x+3)
х1= -5

Answer 1

$y'= (frac{2x-1}{4x+3}) '= frac{(2x-1)'(4x+3)-(4x+3)'(2x+1)}{ (4x+3)^{2} }= frac{2(4x+3)-4(2x+1)}{(4x+3)^{2}} \ frac{8x+6-8x-4}{(4x+3)^{2}} = frac{2}{(4x+3)^{2}} \ x=-5$ 
$frac{2}{(4*(-5)+3) ^{2} } = frac{2}{289}$