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.

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

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

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

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

How do you implicitly differentiate 1=x^2-y+y^21=x^2-y+y^2?