Question

Решите пожалуйста во вложениях найдите производную функцию

Answer 1

$f(x)=(2x-3) sqrt{x} \ \ f'(x)=(2x-3)'sqrt{x}+(2x-3)(sqrt{x})'=(2x-3)'sqrt{x}+(2x-3)(x^{frac{1}{2}})'= \ \ =2sqrt{x}+(2x-3)*frac{1}{2}x^{(-frac{1}{2})}=2sqrt{x}+(2x-3)*frac{1}{2}*frac{1}{sqrt{x}}=2sqrt{x}+frac{2x-3}{2sqrt{x}}= \ \ = frac{2sqrt{x}*2sqrt{x}+2x-3}{2sqrt{x}} = frac{4x+2x-3}{2sqrt{x}} =frac{6x-3}{2sqrt{x}}=frac{3(2x-1)}{2sqrt{x}} \ \ f'(3)=frac{3(2*3-1)}{2sqrt{3}}=frac{3*5}{2sqrt{3}}=frac{15}{2sqrt{3}}=4,3301$

$f(x)=x*sin(x) \ \ f'(x)=(x)'*sin(x)+x*(sin(x))'=x^0*sin(x)+x*cos(x)= \ \ =1*sin(x)+x*cos(x)=sin(x)+x*cos(x) \ \ f'(3)=sin(3)+3*cos(3)=3,0482$

$f(x)=frac{x^3}{6}-0,5x^2-3x+2=frac{1}{6}x^3-frac{1}{2}x^2-3x+2 \ \ f'(x)=frac{1}{6}(x^3)'-frac{1}{2}(x^2)'-3(x)'+(2)'= \ \ =frac{1}{6}(3x^2)-frac{1}{2}(2x)-3(x^0)+(0)=frac{x^2}{2}-x-3= frac{x^2-2x-6}{2} \ \ f'(3)=frac{3^2-2*3-6}{2}=frac{9-6-6}{2}=-frac{3}{2}=-1,5$