# Determine dy/dx if F(x,y)=0 defines y like a function of x: F(x,y)=x^3-2x^2(y)+y-1 I don't understand what this question is asking at all

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You need to substitute `x^3-2x^2(y)+y-1` for F(x,y) in equation F(x,y) = 0 such that:

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

The problem provides the information that you need to use the equation `x^3-2x^2(y)+y-1 = 0` to express y like a function of x, hence you need to isolate the terms containing y to the left side such that:

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

You need to factor out y to the left side such that:

`y(1 - 2x^2) = 1 - x^3`

You need to divide by `1 - 2x^2` both sides such that:

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

You need to differentiate y with respect to x using quotient formula such that:

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

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

`dy/dx = (-3x^2 + 6x^4 + 4x - 4x^4)/((1 - 2x^2)^2)`

Collecting like terms yields:

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

**Hence, using the equation `F(x,y)=0 ` to write y like a function of x yields `dy/dx = (2x^4 - 3x^2 + 4x)/((1 - 2x^2)^2)` **