How do you implicitly differentiate #xy^2 + xy = 12#?

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Answer 1 · 2015-12-24T05:46:05

#dy/dx = (-y^2 -y)/(2xy+x) = (-y(y+1))/(x(2y+1))#

Given #xy^2 + xy = 12#

We can differentiate with respect to #d/dx# like this

#d/dx(xy^2) + d/dx(xy) = d/dx(12)#

This can be done by product rule like this

#d/dx(x) (y^2) + x*d/dx(y^2) + d/dx(x)*y + x*d/dx(y) = d/dx(12)#

#=> y^2 + 2xy dy/dx + y + x(dy/dx) = 0 #

Keep all #dy/dx# on one side like this

#2xy (dy/dx) + x (dy/dx) = -y^2 -y#

Factor out #dy/dx#

#dy/dx(2xy +x) = -y^2 -y#

Divide both side by #2xy + x# to get #dy/dx# by itself

#(dy/dx)(2xy +x)/(2xy+x )=( -y^2 -y)/(2xy+x)#

#dy/dx = (-y^2 -y)/(2xy+x) = (-y(y+1))/(x(2y+1))#