Question

How do you differentiate `1/cos(x) = x/y^2-y`?

Answer 1

Use implicit differentiation and algebra to get `dy/dx=(y(1-sec(x)tan(x)y^2) )/(2x+y^3)`

Explanation:

Assume that the given equation, which is equivalent to `sec(x)=x/y^2-y`, implicitly defines `y` as a function of `x` (the assumption is true wherever the graph of this equation does not have a vertical tangent line).

Now differentiate both sides with respect to `x`, keeping in mind the assumption we made while also using the quotient rule and chain rule:

`sec(x)tan(x)=(y^2-2xy * dy/dx)/y^4-dy/dx`.

This simplifies to

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

Now solve for `dy/dx` to get

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

`=(1-sec(x)tan(x)y^2)/y^2 * y^3/(2x+y^3)`

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