2*cos²2x=1+2*sin²2x
2*cos²x-2*sin²2x=1
2*cos4x=1
cos4x=1/2
4x=+/- π/3+2πn
x=+/-π/12+2πn ⇒
x₁=π/12+2πn
x₂=-π/12+2πn.
Ответом будет пункты 1 и 4
1) y'= 3^x * ln3 -2
2)y'=5^6x-1 * ln5*(6x-1)' +e^3x * (3x)' = 5^6x-1 * 6ln5 + 3e^3x
<span>3)y=log3*(x^2+2x+4)
y'= 1/[(x</span>²+2x+4)*ln3] *(2x+2)
<span>
4)y=ln*(x^2-3x)+cos3x
y'= 1/(x</span><span>²-3x) * (2x-3) -sin3x * 3</span>
(x^2+2x-35)/(25-x^2)=((x+7)(x-5))/((5-x)(5+x))=-(x+7)/(5+x)