How do you differentiate #(y-1)^2+x=0#?

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Answer 1 · 2016-10-16T20:37:31

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

Letting #v = u^2# and #u = y - 1#.

#dy/dx = 2u xx 1(dy/dx) = 2(y- 1)(dy/dx)#

We can re-insert this into the initial relation:

#2(y - 1)(dy/dx) + 1 = 0#

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

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

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

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

Hopefully this helps!