Question

Найдите первую и вторую производной функции 1) y=x^2+5x+8 2)f(x)=2^x+x^2 3)f(x)=(x+1)(3x^2-2x+1) 4)f(x)=cos3x/x^2

Answer 1

$1) y'=(x^2+5x+8)'=2x+5; y''=(2x+5)'=2\\ 2) f'(x)=(2^x+x^2)'=ln2*2^x+2x; f''(x)=(ln2*2^x+2x)'=ln^{2}2*2^x+2\\ 3) f'(x)=((x+1)(3x^2-2x+1))'=(3x^2-2x+1+(x+1)*(6x-2))=9x^2+2x-1; f''(x)=(9x^2+2x-1)'=18x+2\\ 4) f'(x)=(\frac{cos3x}{x^2})'= (\frac{-3sin3x*x^2-2x*cos3x}{x^4})=-(\frac{3sin3x*x+2cos3x}{x^3});\\ f''(x)=(-(\frac{3sin3x*x+2cos3x}{x^3}))'=-(\frac{(9cos3x*x+3sin3x-6sin3x)*x^3-3x^2*(3sin3x*x+2cos3x)}{x^6})=\\-(\frac{9cos3x*x^4-3sin3x*x^3-9sin3x*x^3-6cos3x*x^2}{x^6})=$

$=\frac{12sin3x*x^3-9cos3x*x^4+6cos3x*x^2}{x^6}=\frac{12sin3x*x-9cos3x*x^2+6cos3x}{x^4}$