For $\small{f(x) = \sqrt{x+3}}$, find $\small{f\;'(x)}$ using the definition $\small{f\;'(x) = \displaystyle \lim_{h\to 0}} \large{\frac{f(x + h) - f(x)}{h}}$.

$\small{f\;'(x) =}$

Using this, find the tangent line to the graph of $\small{y = \sqrt{x+3}}$ at $\small{x = 1}$. Write the equation of the line in slope-intercept form.

$\small{y =}$