Question

Найти f‘(x0), f(x)=4x+1/x+3, x0=-2

Answer 1

$(frac{4x+1}{x+3})'= frac{(4x+1)'(x+3)-(4x+1)(x+3)'}{(x+3)^2}= frac{4(x+3)-(4x+1)}{x^2+6x+9}= frac{4x+12-4x-1}{x^2+6x+9}= \ frac{11}{x^2+6x+9} \ \ x_0=-2 \ \ frac{11}{(-2)^2+6(-2)+9}= frac{11}{4-12+9}=11$

Answer 2

f`(x)=[(4x+1)`(x+3)-(4x+1)(x+3)`]/(x+3)²=[4*(x+3)-1*(4x+1)]/(x+3)²=(4x+12-4x-1)/(x+3)²=
=11/(x+3)²
f`(-2)=11/(-2+3)²=11/1=11