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

1 Answer
Sep 26, 2016

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)#