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

Question

1330 views around the world

You can reuse this answer
Creative Commons License

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))#