Question

Найдите производное функций

Answer 1

$195.1)\; \; f(x)=cosx(cosx-1)=cos^2x-cosx\\\\f'(x)=2cosx\, (-sinx)+sinx=-sin2x+sinx\\\\2)\; \; f(x)=tgx(cosx+2)=sinx+2tgx\\\\f'(x)=cosx+\frac{2}{cos^2x}\\\\3)\; \; f(x)=sinx(ctgx-1)=cosx-sinx\\\\f'(x)=-sinx-cosx\\\\4)\; \; f(x)=(4x-1)sinx\\\\f'(x)=4\, sinx+(4x-1)cosx\\\\196.1)\; \; f(x)=cos^2x-1\\\\f'(x)=2cosx\, (-sinx)-0=-sin2x\\\\2)\; \; f(x)=3sin^22x+2x\\\\f'(x)=3\cdot 2sin2x\cdot cos2x\cdot 2+2=6sin2x+2\\\\3)\; \; f(x)=(sin2x+1)^2\\\\f'(x)=2(sin2x+1)\cdot cos2x\cdot 2=4(sin2x+1)cos2x$

$4)\; \; f(x)=(cos2x+sin2x)^3\\\\f'(x)=3(cos2x+sin2x)^2\cdot (-2sin2x+2cos2x)\\\\197.1)\; \; f(x)=\frac{3x+4}{cosx}\\\\f'(x)=\frac{3\cdot cosx-(3x+4)(-sinx)}{cos^2x}=\frac{3\cdot cosx+(3x+4)\cdot sinx}{cos^2x}\\\\2)\; \; f(x)=\frac{5x-2}{sinx}\\\\f'(x)=\frac{5\cdot sinx-(5x-2)\cdot cosx}{sin^2x}\\\\3)\; \; f(x)=\frac{tgx}{3+x}\\\\f'(x)=\frac{\frac{1}{cos^2x}\cdot (3+x)-tgx}{(3+x)^2}=\frac{3+x-sinx\cdot cosx}{cos^2x\cdot (3+x)^2}\\\\4)\; \; f(x)=\frac{x-2}{ctgx}\\\\f'(x)=\frac{ctgx-(x-2)\cdot (-\frac{1}{sin^2x})}{ctg^2x}=\frac{sinx\cdot cosx+x-2}{sin^2x\cdot ctg^2x}$