А) x^3 + y^3 - 6xy = 0
Производная неявно заданной функции.
3x^2 + 3y^2*y' - 6y - 6x*y' = 0
Делим всё на 3
x^2 + y^2*y' - 2y - 2x*y' = 0
y'*(y^2 - 2x) = - x^2 + 2y
y' = (-x^2 + 2y) / (y^2 - 2x)
Б) y = (sin x)^(5x/2)
Производная такой функции равна сумме производных от степенной и от показательной функции.
y = f(x)^g(x)
y' = g*f^(g-1) *f'(x) + f^g*ln |f|*g'(x)
В нашем случае f = sin x; g = 5x/2.
y' = (5x/2)*(sin x)^(3x/2)*(cos x) + (sin x)^(5x/2)*(ln |sin x|)*5/2
В) x = √(2t - t^2); y = (1-t)^(-2/3)
y'(x) = dy/dx = (dy/dt) : (dx/dt)
dx/dt = (-2t+2) / [2√(2t-t^2)] = (-t+1) / √(2t-t^2)
dy/dt = -(-2/3)*(1-t)^(-5/3) = (2/3) / (1-t)^(5/3)
dy/dx = [(2/3) / (1-t)^(5/3)] : [(-t+1) / √(2t-t^2)] =
= [(2/3)*√(2t-t^2)] / [(1-t)^(5/3)*(1-t)] = [2/3*√(2t-t^2)] / [(1-t)^(8/3)]
2) tg (x/2 + п/4) = -1⇒x/2+π/4=-π/4+πn⇒x/2=-π/2+πn⇒x=-π+2πn
3) sin 2x + cos x = 0⇒2sinxcosx+cosx=0⇒cosx(2sinx+1)=0
cosx=0⇒x=π/2+πn U sinx=-1/2⇒x=(-1)^n+1 *π/6+πn
4) cos 7x + cos x = 0⇒2cos4xcos3x=0
cos4x=0⇒4x=π/2+πn⇒x=π/8+πn/4 U cos3x=0⇒3x=π/2+πn⇒x=π/6+πn/3
5) 2cos²x + 2sinx = 2,5
2(1-sin²x)+ 2sinx = 2,5
2sin²x-2sinx+0,5=0
sinx=a
2a²-2a+0,5=0
4a²-4a+1=0
(2a-1)²=0
a=1/2⇒sinx=1/2⇒x=(-1)^n *π/6+πn
6) 1 + 2sin 2x + 2cos²x = 0
cos²x+sin²x+4sinxcosx+2cos²x=0
sin²x+4sinxcosx+3cos²x=0 /cos²x≠0
tg²x+4tgx+3=0
tgx=a
a²+4a+3=0
a1+a2=-4 U a1*a2=3
a1=-3⇒tgx=-3⇒x=-arctg3+πn
a2=-1⇒tgx=-1⇒x=-π/4+πn