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How do you find #dy/dx# by implicit differentiation of #x^2+y^2=36#?

1 Answer

Jan 20, 2017

#dy/dx = -y/x#

Explanation:

#d/dx(x^2+y^2) = d/dx36#

#=> d/dxx^2 + d/dxy^2 = 0#

#=> 2x + 2y(d/dxy) = 0#
(Applying the chain rule with #y# treated as a function of #x#)

#=> 2x+2ydy/dx = 0#

#=> 2ydy/dx = -2x#

#:. dy/dx = -x/y#

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