Question

How do you implicitly differentiate `x^(2/3) + y^(2/3) = 4`?

Answer 1

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

Explanation:

Differentiating with respect to x :

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

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

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

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

now cancelling `2/3` and writing with positive indices gives :

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