1) y=2sin(4x)-8cos(x/4)+(1/2)*tg(2x)-(1/12)*ctg(6x)
y ' =8cos(4x)+2sin(x/4)+1/cos^2(x)+sin^2(x)/2
2) y=sin(x/4)+12cos(x/3)-10tg(x/2)+5ctg(2x)
y ' = cos(x/4)/4-sin(x/3)/3-5/cos^2(x/2)+2*sin^2(2x)/5
3) y=(8/12)*sin(3x/4)-(4/3)*cos(3x/4)-40ctg(x/5)-tg(8x)
y ' = (1/2)*sin(3x/4)+sin(3x/4)+8/sin^2(x)-8/cos^2(x)
4) y =cos(2x)*x^5
y ' =-2sin(2x)*x^5+5cos(2x)*x^4
5) y=sin(2x)/cos(4x)
y ' =2cos(2x)/cos(4x)+4sin(2x)/cos^2(4x)
6) y=8cos(4x-pi/3)
y ' =-32sin(4x-pi/3)
7) y=10x^5+7x^4-8x^3+4/x-9sqrt(x)-4x+1,1
y ' = 50x^4+28x^3-24x^2-4/x^2-9/2*sqrt(x)-4
8) y=sin(3x)*tg(3x)
y ' = 3cos(3x)*tg(3x)+sin(3x)*3/cos^2(3x)
9) y=5x^6+2x^3+6x^2-6x-8
y ' = 30x^5+6x^2+12x-6
y '' = 150x^4+12x+12
10) y=4sin(2x)-16cos(4/x)
y ' = 8cos(2x)+64sin(x/4)/x^2
y '' =-16sin(2x) +16cos(x/4)/x^2-128sin(x/4)/x^3
Докажите тождество:
1. (3−4cos2α+cos4α)/(3+4cos2α+cos4α) =tg⁴α
2. (1+sin2α+cos2α)/(1+sin 2α−cos2α)=ctg α
Объяснение:
* * *сos2φ = cos²φ- sin²φ =2cos²φ - 1 = 1 - 2sin²φ * * *
1. (3−4cos2α+cos4α)/(3+4cos2α+cos4α) =
(3−4cos2α+2cos²2α -1 )/ (3+4cos2α+2cos²2α-1)=
2(1 - 2cos2α+cos²2α) / 2( 1 +2cos2α+cos²2α) =
(1-cos2α)² / (1+cos2α)²= 4sin⁴α /4cos⁴α =tg⁴α ч.т.д.
* * * sin2α =2sinα*cosα * * *
2. (1+sin2α+cos2α) / (1+sin 2α−cos2α)=
(2cos²α+2sinα*cosα) / (2sin²α+2sinα*cosα ) =
2( cosα+sinα)*cosα) / 2(sinα+cosα )*sinα =cosα) / sinα =ctgα.