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How do you find dy/dx of #y^2 = ln x# and evaluate it at the point (e, 1)?

1 Answer

Apr 2, 2018

#1/[2e#

Explanation:

#y^2=lnx#. Need to differentiate both sides of this expression implicitly with respect to #x#

#d/dx[y^2=lnx]#=#[d/dxy^2=d/dxlnx]#.

#d/dx y^2=2ydy/dx# and #d/dxlnx=1/x# and so ,

#2ydy/dx=1/x#, i.e, #dy/dx=1/[2yx#.....#[1]#

substituting #x=e and y=1# into .....#[1]# gives #dy/dx=1/[2e#.
Hope this helps.

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