Question

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

Answer 1

`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)`

Answer 2

`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)`