# What is the equation of the tangent line of f(x)=sqrt(x+1)  at x=0?

Jun 13, 2018

$2 y - x - 2 = 0$

#### Explanation:

We have $f \left(0\right) = 1$, and

${f}^{'} \left(x\right) = \frac{1}{2 \sqrt{x + 1}}$

and thus

${f}^{'} \left(0\right) = \frac{1}{2}$

So, the required tangent is a straight line with slope $\frac{1}{2}$, passing through $\left(0 , 1\right)$

It has the equation

$y - 1 = \frac{1}{2} \left(x - 0\right) \implies 2 y - x - 2 = 0$