Find the slope of the tangent line to the graph of x= 5y^3-6y ^2 - 3 at (-4,1)

Implicit differentiation

take derivitive of both sides

1=15y^2dy/dx-12ydy/dx
1=dy/dx(15y^2-12y)
divide both sides by (15y^2-12y)
1/(15y^2-12y)=dy/dx

sub the point (-4,1)
1/(15(1)^2-12(1))=dy/dx
1/(15-12)=dy/dx
1/3=dy/dx
slope is 1/3

Z1 = 5(cos25˚+isin25˚)

Z2  = 2(cos80˚+isin80˚)

Z1.Z2  = 5(cos25˚+isin25˚). 2(cos80˚+isin80˚)

Z1.Z2  = 10{(cos25˚cos80˚ + isin25˚cos80˚+i^2sin25˚sin80˚) }

Z1.Z2  =10{(cos25˚cos80˚- sin25˚sin80˚+ i(cos25˚sin80˚+sin25˚cos80˚))}

(i^2 = -1)

Cos(A+B) = cosAcosB – sinAsinB

Sin(A+B) = sinAcosB + cosAsinB

Z1.Z2 = 10(cos(25˚+80˚) +isin(25˚+80˚)

Z1.Z2     = 10(cos105˚+ isin105˚)