Question

Надо! вычислить производную тригонометрической функции: f(x)=tg^2 3x , при х= pi/12 f(x)=tg^2 x * sinx , при х= pi/3 f(x)=8*sin^2 x * cosx , при х= pi/4 f(x)=tg^2 x - ctg^2 x , при х= pi/4 f(x)=cos 2x * (1+sin 2x) , при х= pi/8 f(x)=1-tg 2x / tg 2x , при х= pi/4

Answer 1

1. f '(x)=(tg²3x)'=2tg3x*(tg3x)'=2tg3x*(1/cos²3x)*(3x)'=6tg3x/cos²3x
f '(π/12)=6tg(3*π/12)/cos²(3*π/12)=6*tg(π/4)/cos²(π/4)=6*1/(√2/2)²=12

2. f'(x)=(tg²x)'=2tgx*(tgx)'=2tgx*(1/cos²x)=2tgx/cos²x
f'(π/3)=2tg(π/3)/cos²(π/3)=2*√3/(√3/2)²=3/2=1,5

3. f(x)=8sin²x*cosx, f(x)=8(1-cos²x)cosx=8cosx-8cos³x
f'(x)=(8cosx-8cos³x)'=-8sinx-8*3cos²x*(cosx)'=-8sinx-24cos²x*(-sinx)=-8sinx*(1-cos²x)=-8sinx*sin²x=-8sin³x
f'(π/4)=-8*(√2/2)²=-2√2

4. f'(x)=(tg²x-ctg²x)'=2tgx*(tgx)'-2ctgx*(ctgx)'=2tgx/(1/cos²x)-2ctgx*(-1/sin²x)=2tgx/cos²x+2ctgx/sin²x
f'(π/4)=2tg(π/4)/(cosπ/4)²+2ctg(π/4)/(sinπ/4)²=2*1/(√2/2)²+2*1/(√2/2)²=8

5. f'(x)=(cos2x(1+sin2x))'=-sin2x*(2x)' *(1+sin2x)+cos2x*cos2x*(2x)'=-2sin2x-2sin²2x+2cos²2x=2(cos²2x-sin²2x)=2cos4x
f'(π/8)=2cos(4*π/8)=2cosπ/2=2*0=0

6. f'(x)=((1-tg2x)/tg2x)'=[-(1/cos²2x)*(2x)' *tg2x-(1-tg2x)*(1/cos²2x)*(2x)'] / (tg2x)²=[-2tg2x/cos²2x-2(1-tg2x)/cos²2x ] /tg²2x=-2/(cos²2x*tg²2x)=-2/sin²2x
f'(π/4)=-2/sin²(2*π/4)=-2/(sinπ/2)²=-2/1=-2