Question

How do you determine dy/dx given `x^(1/3)+y^(1/3)=7`?

Answer 1

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

Explanation:

Note that `d/dxx^(1/3)=1/3x^(-2/3)` through the power rule.

The derivative with respect to `x` of `y^(1/3)` is found identically, except for that since `y` is its own function, the chain rule will come into effect, and we will have to multiply the derivative found through the product rule by the derivative of `y`, which is `dy/dx`.

That is `d/dxy^(1/3)=1/3y^(-2/3)*dy/dx`.

So we have:

`x^(1/3)+y^(1/3)=7`

And taking the derivative of both sides:

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

Multiplying both sides by `3` and rewriting:

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

Solving for `dy/dx`:

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

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