A curve C has the following equation: x^3 + 3y - 4(x^3)*(y^3) a) Show that (1,1) lies on C b) Find dy/dx

a) Substituting the coordinate (1,1) into the left hand side of the equation for C we obtain: (13) + 3*1 - 4(13)(13) = 1 + 3 - 4 = 0 = The right hand side of the equation, hence the equation is satisfied, and therefore (1,1) lies on C.

b) Differentiating implicity we find: 
3x2 + 3dy/dx - 12x2y- 12x3y2dy/dx = 0
Rearranging yields:
3x2 - 12x2y3= (12x3y2 - 3)dy/dx

Hence dy/dx = (3x2 - 12x2y3/(12x3y2 - 3)
Which simplifies to 

dy/dx = x2(1 - 4y3)/(4x3y2 - 1)

(An alternative expression can be obtained be moving the terms not involving dy/dx to the right hand side)

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Answered by Harry W. Maths tutor

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