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

Apr 2, 2018

1/[2e

#### Explanation:

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

$\frac{d}{\mathrm{dx}} \left[{y}^{2} = \ln x\right]$=$\left[\frac{d}{\mathrm{dx}} {y}^{2} = \frac{d}{\mathrm{dx}} \ln x\right]$.

$\frac{d}{\mathrm{dx}} {y}^{2} = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$ and $\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$ and so ,

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

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