Answer
$$y=x$$
Work Step by Step
We have: $y= \ln (\sqrt {2x+1})$
We differentiate both sides with respect to $x$.
$y^{\prime}=\dfrac{1}{ \sqrt {2x+1})} \times \dfrac{1}{2\sqrt {2x+1}} \times (2)\\=\dfrac{1}{2x+1}$
Now, $y^{\prime}|_{x=0}=1$
The equation of a line is:
$$\dfrac{y-0}{x-0}=1\\\implies y=x$$
