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

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Answer 1 · 2018-04-02T20:08:53

$\frac{1}{2 e}$ $1/[2e$

$y^{2} = ln x$ $y^2=lnx$ . Need to differentiate both sides of this expression implicitly with respect to $x$ $x$

$\frac{d}{d x} [y^{2} = ln x]$ $d/dx[y^2=lnx]$ = $[\frac{d}{d x} y^{2} = \frac{d}{d x} ln x]$ $[d/dxy^2=d/dxlnx]$ .

$\frac{d}{d x} y^{2} = 2 y \frac{d y}{d x}$ $d/dx y^2=2ydy/dx$ and $\frac{d}{d x} ln x = \frac{1}{x}$ $d/dxlnx=1/x$ and so ,

$2 y \frac{d y}{d x} = \frac{1}{x}$ $2ydy/dx=1/x$ , i.e, $\frac{d y}{d x} = \frac{1}{2 y x}$ $dy/dx=1/[2yx$ ..... $[1]$ $[1]$

substituting $x = e and y = 1$ $x=e and y=1$ into ..... $[1]$ $[1]$ gives $\frac{d y}{d x} = \frac{1}{2 e}$ $dy/dx=1/[2e$ .
Hope this helps.

How do you find dy/dx of y^2 = ln xy2=lnxy^2 = ln x and evaluate it at the point (e, 1)?