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

Question

1319 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-21T18:03:11

You can do it like this:

#x-y^2=ysin(x)+y#

Differentiate both sides implicitly with respect to #x#:

#D(x-y^2)=D[ysin(x)+y]#

#:.1-2y.y'=ycos(x)+sin(x).y'+y'#

#:.1-ycos(x)=2y.y'+sin(x).y'+y'#

#:.1-ycos(x)=y'(2y+sin(x)+1)#

#:.y'=((1-ycos(x)))/((2y+sin(x)+1))#