How do you determine dy/dx given x^(2/3)+y^(2/3)=a^(2/3)?

2 Answers
Nov 14, 2016

dy/dx=-(y/x)^(1/3)

Explanation:

We will differentiate implicitly. This means, that whenever we differentiate a variable that's not x, the chain rule will kick in. Take a look at a couple examples before starting here:

  • d/dxx^3=3x^2" " but " "d/dxy^3=3y^2dy/dx
  • d/dxx=1" " but " "d/dxy=dy/dx
  • d/dxxy^2=(d/dxx)y^2+x(d/dxy^2)=y^2+x(2y)dy/dx

So, for the given example, we see that we will use the power rule on the left-hand side. On the right, don't be misled by the two-thirds power: a^(2/3) is still a constant.

d/dx(x^(2/3)+y^(2/3))=d/dxa^(2/3)

Power rule:

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

Multiply both sides by 3/2:

x^(-1/3)+y^(-1/3)dy/dx=0

1/x^(1/3)+1/y^(1/3)dy/dx=0

Solving for dy/dx:

1/y^(1/3)dy/dx=-1/x^(1/3)

dy/dx=-y^(1/3)/x^(1/3)=-(y/x)^(1/3)=-root3(y/x)

Nov 14, 2016

dy/dx=-root2((a^(2/3)-x^(2/3)))/root3(x)

Explanation:

Rewrite it y^(2/3)=a^(2/3)-x^(2/3) so that we have
y=root2((a^(2/3)-x^(2/3))^3) and the derivative is

dy/dx=3/2root2((a^(2/3)-x^(2/3)))*(-2/3x^(-1/3))

dy/dx=-root2((a^(2/3)-x^(2/3)))/root3(x)