What is the derivative of #y^2=x# where #y=1/2#?

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Answer 1 · 2014-12-08T23:42:59

#y'=1#

#y^2=x#

Differentiating both sides implicitly:

#2y.y'=1#

#y'=1/(2y)#

If y=1/2 then:

#y'=1/(2xx1/2)=1#

#y'=1# graph{(y-sqrt(x))(y-x-1/4)=0 [-2.5, 2.5, -1.25, 1.25]}