Question

СРОЧНО!АЛГЕБРА 10 КЛАСС! ДАЮ 80Б.ТОЛЬКО С РЕШЕНИЕМ

Answer 1

1) f'(x) = (x^3)' = 3x^(3-1) = 3x^2
2) f'(x) = (3x)' = 3
3) f'(x) = (x^(-8))' = -8x^(-9)
4) f'(x) = (3,7)' = 0
5) f'(x) = (1/x)' = (x^(-1))' = -x^(-2) = -1/x^2
6) f'(x) = (3x^8-2x^6+4x^4-3x^2+x-8)' =
= 3*8x^(8-1)-2*6x^(6-1)+4*4x^(4-1)-3*2x^(2-1)+x^0-0 =
= 24x^7-12x^5+16x^3-6x+1
7) 2) f'(x) = (x^4+$\sqrt{x}$)' =  (x^4+x^(1/2) )' =
= 4x^(4-1)+1/2 *x^(1/2-1) = 4x^3-1/2*x^-1/2
4x^3-$\frac{1}{2 \sqrt{x} }$
8) f'(x) = (1/4 *x^(-4))' = 1/4*(-4)*x^(-4-1) = -x^(-5)
9) 6) f'(x) = (x^8-3x^4-x+5)' = 8x^(8-1)-3*4x^(4-1)-1X^0+0 = 8x^7-12x^3-1

Answer 2

$1)f(x)=x^3\\f'(x)=3x^2\\2)f(x)=3x\\f'(x)=3\\3)x^{-8}\\f'(x)=-\frac{8}{x^9} \\4)f(x)=3,7\\f'(x)=0\\5)f(x)=\frac{1}{x} =\frac{(1)'x-1(x')}{x^2} =-\frac{1}{x^2} \\6)f(x)=3x^8-2x^6+4x^4-3x^2+x-8\\f'(x)=24x^7-12x^5+16x^3-6x+1\\7)f(x)=x^4- \sqrt{x} \\f'(x)=4x^3-\frac{1}{2 \sqrt{x} } =\frac{8x^3 \sqrt{x} -1}{2 \sqrt{x} } \\8)f(x)=\frac{1}{4x^{-4}} =-\frac{4(-4x^{-5})}{(4x^{-4})^2} =x^3\\9)f(x)=x^8-3x^4-x+5\\f'(x)=8x^7-12x^3-1$