Question

Найти значение производной в точке

Answer 1

f'(x)= (x^2-1)'(2x+1)- (x^2-1)(2x+1)'/(2x+1)^2=
=2x(2x+1)-2(x^2-1)/(2x+1)^2=
=4x^2+2x-2x^2+2/(2x+1)^2=
=2x^2+2x+2/(2x+1)^2;
f'(1)=2*1^2+2*1+2/(2*1+2)^2=
=6/8=3/4

Answer 2

$f'(x)=( frac{ x^{2} -1}{2x+1} )'= frac{( x^{2} -1)' *(2x+1)-(2x+1)' *( x^{2} -1)}{ (2x+1)^{2} } = frac{2x*(2x+1)-2*( x^{2} -1)}{(2x+1)^{2} } =$
$= frac{4 x^{2} +2x-2 x^{2} +2}{ (2x+1)^{2} } = frac{2 x^{2} +2x+2}{(2x+1) ^{2} }$
$f'(1)= frac{2* 1^{2}+2*1+2 }{(2*1+1) ^{2} } = frac{6}{9} = frac{2}{3}$