Question

Find dy/dx if x^3 + y^3 -3axy=0

Answer 1

Required" dy/dx of x^3 + y^3 -3axy = 0"

we know that

[d/dx (c) = 0 (where c is a constant)

d/dx (x) = 1

d/dx (x^n) = n.x^(n-1)

d/dx (u.v) = u.d/dx .v + v.d/dx .u]

Apply the above formulas...

x^3 + y^3 - 3axy = 0

d/dx (x^3) + d/dx (y^3) - d/dx (3axy) = 0

3.x^(3-1) + 3.y^(3-1)d/dx -3a (x.d/dx y + y.d/dx x) = 0

3x^2 + 3y^2 d/dx - 3a (x.d/dx y + y .(1)) = 0

3x^2 + 3y^2 d/dx - 3a (x.dy/dx + y .(1)) = 0

3x^2 + 3y^2 dy/dx - 3ax.dy/dx - 3ay = 0

3x^2 + dy/dx( 3y^2 - 3ax) -3ay = 0

dy/dx(3y^2 - 3ax) = 3ay - 3x^2

dy/dx =( 3ay - 3x^2) /(3y^2 -3ax) is the required answer..

^^^^^Thank You^^^^^