Question

найти значение производной функции f (x) в точке, где абсцисса x0

Answer 1

1)f`(x)=x-3+x=2x-3

f`(4)=8-3=5

2)f`(x)=2x(x-3)+(x^2-5)*1=2x^2-6x+x^2-5=3x^2-6x-5

f`(1.1)=3*1.21-6.6-5=-7.97

3)f`(x)=4((2x+3)(x-1)+x^2+3x)=4(2x^2+x-3+x^2+3x)=4(3x^2+4x-3)=12x^2+16x-12

f`(-0.4)=12*0.16-16*0.4-12=1.92-6.4-12=-16.48

4)f`(x)=2(x+3)+2x-1-1=4x+4

f`(4/3)=4*4/3+4=5 1/3+4=9 1/3

Answer 2

$1)\ \ f(x)=x\cdot (x-3)\ \ ,\ \ \ \ f(x)=x^2-3x\\\\f'(x)=2x-3\ \ ,\ \ \ \ f'(4)=2\cdot 4-3=8-3=5\\\\\\2)\ \ f(x)=(x^2-5)(x-3)\\\\f'(x)=2x\cdot (x-3)+(x^2-5)\cdot 1=2x^2-6x+x^2-5=3x^2-6x-5\\\\f'(1,1)=3\cdot 1,21-6\cdot 1,1-5=-7,97\\\\\\3)\ \ f(x)=4\cdot (x^2+3x)\cdot (x-1)\ \ ,\ \ \ f(x)=(4x^2+12x)\cdot (x-1)\\\\f'(x)=(8x+12)\cdot (x-1)+(4x^2+12x)\cdot 1=8x^2+4x-12+4x^2+12x=\\\\=12x^2+16x-12\\\\f'(-0,4)=12\cdot 0,16-16\cdot 0,4-12=-16,48$

$4)\ \ f(x)=(2x-1)(x+3)-x\\\\f'(x)=2(x+3)+(2x-1)\cdot 1-1=2x+6+2x-1-1=4x+4\\\\f'\Big(1\dfrac{1}{3}\Big)=f'\Big(\dfrac{4}{3}\Big)=4\cdot \dfrac{4}{3}+4=\dfrac{16+12}{3}=\dfrac{28}{3}=9\dfrac{1}{3}$