Question

How do you find `(dy)/(dx)` given `root3x+root3(y^4)=2`?

Answer 1

Start by writing the roots as rational exponents:

`x^(1/3) + y^(4/3) = 2`

Differentiate each term using the power rule:

`1/3x^(1/3 - 1) + 4/3y^(4/3 - 1)dy/dx = d/dx(2)`

`1/3x^(-2/3) + 4/3y^(1/3) dy/dx = 0`

`1/(3x^(2/3)) + 4/3y^(1/3)dy/dx = 0`

`4/3y^(1/3)dy/dx = 0- 1/(3x^(2/3))`

`dy/dx = (- 1/(3x^(2/3)))/(4/3y^(1/3))`

`dy/dx = ( - 1)/(3x^(2/3)) xx 1/(4/3y^(1/3))`

`dy/dx = - 1/(4x^(2/3)y^(1/3))`

If you want it in radical form instead of rational exponent form:

`dy/dx = - 1/(4root(3)(x^2)root(3)(y))`

Hopefully this helps!