Question

1 . Найдите значение производной функции в точке х0
f(х)=3х^2+4х-1 ; х0=-2
2. y=2x^3/3x-1

Answer 1

$f'=6x+4$
$f'(-2)=6*(-2)+4=-8$

$f'= \frac{6 x^{2} *(3x-1)-3*2x^3}{(3x-1)^2} = \frac{12x^3-6 x^{2} }{(3x-1)^2}$
$f'(-2)= \frac{12*(-8)-6*4}{((-6-1)^2}=- \frac{120}{49} =-2 \frac{22}{49}$

Answer 2

1) $y'=(3x^2 +4x-1)' = 6x+4; \\ y'(x_0)=6*(-2) +4 =-8.$
2)$y'=(2x^3/(3x-1))= \frac{6x^2(3x-1)-6x^3}{(3x-1)^2} = \frac{18x^3-6x^2-6x^3}{(3x-1)^2} = \frac{12x^3-6x^2}{(3x-1)^2} ; \\ y'(x_0) = \frac{12*(-8) - 6*4}{49} = \frac{-96-24}{49} = -\frac{120}{49} =-2 \frac{22}{49} .$