How do you find #dy/dx# given #x=(y^2+2)^-1#?

Question

1014 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-10T18:12:17

Multiply both sides by #(y^2 + 2)#:

#x(y^2 + 2) = 1#

Distribute the x:

#xy^2 + 2x = 1#

#y^2 + 2xy(dy/dx) + 2 = 0#

#2xy(dy/dx) = -2 - y^2#

#dy/dx = -(y^2 + 2)/(2xy)#