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

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Answer 1 · 2016-08-02T15:44:17

# dy/dx = pm 1/ (2 sqrt( x) )#

#y^2 = x#

#d/dx (y^2) =d/dx (x)#

#2y dy/dx =1#

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

and as #y = pm sqrt x#

# dy/dx = pm 1/ (2 sqrt( x) )#

OR

#x = y^2#

#dx/dy = 2y = pm 2 sqrt x#

#dy/dx = 1/(2y)# and same story