<span>1) f`(x)=(5x³-4x²)`=15x²-8x
f`(2)=15·4-8·2=44
2) f`(x)=(2sinx+cosx-ctgx)`=2(sinx)`+(cosx)`-(ctgx)`=
= 2cox-sinx+(1/sin²x)
f`(π/6)=2·cos(π/6)-sin(π/6)+(1/sin²(π/6))=(2√3/2)- (1/2)+(1/(1/4))=√3-0,5+4=3,5+√3
3) f`(x)=(3(2x-1)⁵¹)`=3·(2x-1)⁵⁰·(2x-1)`=6·(2x-1)⁵⁰
f`(2)=6·(2·2-1)⁵⁰=6·3⁵⁰
4) f``(x)=(√(2x²+1))`=(1/2√(2х²+1))·(2х²+1)`=4x/2√(2х²+1)=2х/√(2х²+1)
f`(7)=14/√99
5) f`(x)=(sinx+cosx/sinx-cosx)`=(sinx+cox)`·(sinx-cosx)-(sinx+cosx)·(sinx-cosx)`/(sinx-cosx)²=
=(cosx-sinx)(sinx-cosx)-(sinx+cosx)(cosx+sinx)</span><span>/(sinx-cosx)²=
=-4(sin²x+cos²x)/</span><span>(sinx-cosx)²=-4/</span><span><span><span>(sinx-cosx)²</span>
f(</span>п/2)=-4/(1-0)²=-4
6) f`(x)=(4cos²2x)`=8cos2x·(cos2x)`=8cos2x·(-sin2x)·(2x)`=-8sin4x
f`(π/6)=-8sin(2π/3)=-8sin(π/3)=-4√3</span>
1)=m(n-k)-x(n-k)=(m-x)(n-k)
2)=x(x+7)-a(x+7)=(x-a)(x+7)
3)=m(3-k)+k(3-k)=(m+k)(3-k)
4)=x(k-x)+y(k-x)=(x+y)(k-x)
3 sin x=0
x=пик
4 sin ( пи/2+х)=0
пи/2+х=0
пи/2+х=ПИн
х=пи/2+пн
2xy² * 0,25x²y⁵ = 0,5x¹⁺²y²⁺⁵ = 0,5x³y⁷
Приравниваем оба уравнения и получаем
8/х=х/2
х²=16
х1=4
х2=-4
у1=8/4=2
у2=8/(-4)=-2
ответ:(4;2);(-4;-2)