Question

How do you implicitly differentiate `x^2/16 + y^2/36 = 1`?

Answer 1

Differentiate each term separately, collect the terms containing the factor `dy/dx` on the left, move other terms to the right, and then divide both sides by the coefficient of `dy/dx`

Explanation:

Term 1:

`(d(x^2/16))/dx = 2x/16 = x/8`

Term 2:

`(d(y^2/36))/dx = (2y/36)(dy/dx) = y/18(dy/dx)`

Term 3;

`(d(1))/dx = 0`

Returning the 3 terms to their place in the equation:

`x/8 + y/18(dy/dx) = 0`

Collect terms that contain `dy/dx` on the left and move any other terms to the right:

`y/18(dy/dx) = -x/8`

Divide by the coefficient of `dy/dx`:

`dy/dx = (-x/8)(18/y)`

`dy/dx = (-9x)/(4y)`