How do you implicitly differentiate #1=x^2-y+y^2#?

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Answer 1 · 2015-11-24T03:09:13

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

Note that I'm assuming we are differentiating implicitly with respect to #x#.

#d/dx[1=x^2-y+y^2]#

Remember that any term that's not an #x# (in this case, #y#) will put chain rule into effect and spit out a #dy/dx# term.

#0=2x-dy/dx+2y(dy)/dx#

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

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

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