Question

How do you find `dy/dx` by implicit differentiation given `2x^2-3y^2=4`?

Answer 1

Take the derivative of both sides with respect to `x`, treating `y` as a function of `x`. Solve for `dy/dx`.

Explanation:

`2x^2-3y^2=4`

`=>"    "d/dx(2x^2-3y^2)"      "=d/dx(4)`

`=>d/dx(2x^2)-d/dx(3y^2)=0`

`=>"      "4x"     "-"   "6y dy/dx"   "=0"      "`(by chain rule)

`=>"                  "-6y dy/dx"   "=-4x`

`=>"                           "dy/dx"   "="  "("-"4x)/("-"6y)"   "="   "(2x)/(3y)`

Therefore, `dy/dx=(2x)/(3y)`.