Question

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

Answer 1

`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.