Найти производные функции№1f(t)=0.5t3+0.6t2+0.8t+8 f '(1)-?№2y=4x 3/4 в степени + 3x 2/3...

Решите задачу:

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

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