Question

How do you use implicit differentiation to find `(dy)/(dx)` given `x^(2/3)+y^(2/3)=1`?

Answer 1

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

Explanation:

Use the power rule and implicit differentiation to find the derivative.

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

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

Since we're differentiating with respect to `x`, we must put `dy/dx` at every y term that we differentiate (after differentiating, of course).

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

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

Isolate `dy/dx` and simplify:

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

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

`dy/dx = (-6y^(1/3))/(6x^(1/3))`

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

Hopefully this helps!

Answer 2

`-y^(1/3)/x^(1/3)`

Explanation:

Define `f(x,y)=x^(2/3)+y^(2/3)-1=0`

we know that

`df = f_x dx + f_y dy = 0`

so

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