Question

Найти производные функции

№1
f(t)=0.5t3+0.6t2+0.8t+8 f '(1)-?

№2
y=4x 3/4 в степени + 3x 2/3 в степени + 4x 1/2 в степени + 3x

Answer 1

1) 3*0,5t^2+2*0,6t+0,8=1,5t^2+1,2t+0,8

2) (3/4)*4x^(-1/4)+(2/3)*3x^(-1/3)+(1/2)*4x^(-1/2)+3
=(3/(x^1/4))+(2/(x^1/3))+(2/(x^1/2))+3

Answer 2

2) (3/4)*4x^(-91/4)+(2/3)*3x^(-1/3)+(1/2)*4x^(-1/2)+3
=(3/(x^1/4))+(2/(x^1/3))+(2/(x^1/2))+3

Answer 3

$1. f(t) = 0.5t^3 + 0.6t^2 + 0.8t + 8\\ left[ (cf(t))' = cf'(t); (x^n)' = nx^{n - 1}; (c)' = 0 right]\\ f'(t) = 0.5*3t^2 + 0.6*2t + 0.8 = 1.5t^2 + 1.2t + 0.8\\ f'(1) = 1.5 + 1.2 + 0.8 = boxed{3.5}$

$2. y = 4x^{frac{3}{4}} + 3x^{frac{2}{3}} + 4x^{frac{1}{2}} + 3x\\ y' = 4*frac{3}{4}*x^{frac{3}{4} - 1} + 3*frac{2}{3}*x^{frac{2}{3} - 1} + 4*frac{1}{2}*x^{frac{1}{2} - 1} + 3x^{1 - 1} = \\ = 3x^{-frac{1}{4}} + 2x^{-frac{1}{3}} + 2x^{-frac{1}{2}} + 3 = boxed{frac{3}{sqrt[4]{x}} + frac{2}{sqrt[3]{x}} + frac{2}{sqrt[2]{x}} + 3}$