Question

Question #3aba9

Answer 1

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

Explanation:

We can proceed implicitly, by differentiating both sides of the equation with respect to `x`:

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

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

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

we can also note that for every `x > 0` there are two possible values of `y`:

`y =+- sqrtx (x-1)`

so we can also express the derivative as:

`dy/dx =+-((x-1)(3x-1))/(2sqrtx (x-1)) = +- (3x-1)/(2sqrtx)`

or:

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

Function:

graph{y^2= x(x-1)^2 [-10, 10, -5, 5]}

Derivative:

graph{y^2 = (3x-1)^2/(4x) [-10, 10, -5, 5]}