How do you implicitly differentiate # 2x+2y = sqrt(x^2+y^2)#?

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Answer 1 · 2016-06-15T17:06:06

#dy/dx = { -1/3 ( sqrt[7]+4), 1/3 ( sqrt[7]-4)}#

depending on de sign of #x#

#f(x,y)=2x+2y - sqrt(x^2+y^2)=0# is an homogeneus function.

Making the substitution

#y = lambda x# we get at

#f(x,lambda x) = 2x(1+lambda)-abs(x) sqrt(1+lambda^2) = 0# or

#x/abs(x)=1/2sqrt(1+lambda^2)/(1+lambda) = pm 1#

Solving for #lambda# we have

#x/abs(x) = 1-> lambda = 1/3 ( sqrt[7]-4)#
#x/abs(x) = -1-> lambda = -1/3 ( sqrt[7]+4)#

so #dy/dx = { -1/3 ( sqrt[7]+4), 1/3 ( sqrt[7]-4)}#

depending on de sign of #x#