How do you implicitly differentiate $(x+y)^2-2xy=x+220$ ?

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Answer 1 · 2015-12-12T20:59:53

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

Take the derivative with respect to $x$ . Remember that any $y$ -term that's differentiated will spit out a $dy/dx$ term thanks to the chain rule.

$d/dx[(x+y)^2-2xy=x+220]$

Find each part:

$d/dx[(x+y)^2]=2(x+y)d/dx[x+y]=(2x+2y)(1+dy/dx)$

$d/dx[2xy]=yd/dx[2x]+2xdy/dx=2y+2xdy/dx$

$d/dx[x]=1$

$d/dx[220]=0$

We get:

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

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

$dy/dx(4x+2y)=1-2x$

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

How do you implicitly differentiate (x+y)^2-2xy=x+220(x+y)^2-2xy=x+220?