A curve has equation y^3+2xy+x^2-5=0. Find dy/dx.

To find the derivative of this, we must differentiate each term with respect to x. This implies d/dx(y^3+2xy+x^2-5=0). We can differentiate each term seperately so d/dx(y^3+2xy+x^2-5=0) is equal to d/dx(y^3) + d/dx(2xy) +d/dx(x^2) - d/dx(5) = 0. Taking each term seperately, d/dx(y^3) = (dy/dx)(d/dy(y^3) = dy/dx(3y^2), d/dx(2xy) = 2y+dy/dx(d/dy(2xy)) = 2y+2x(dy/dx), d/dx(x^2) = 2x, d/dx(5) = 0. Recombining we get dy/dx(3y^2)+2y+dy/dx(2x)+2x=0. Rearranging and factorising gives us dy/dx(3y^2+2x)=-(2x+2y). Dividing by 3y^2+2x then gives us dy/dx = -(2x+2y)/3y^2+2x. ARQ.