Find an equation for the line tangent to the graph of \[ f(x)=\frac{\sqrt{x}}{2 x+5} \] at the point $(1, f(1))$. \[ y= \]

Step 1 ：Find the derivative of the function using the quotient rule and chain rule: \(f'(x) = \frac{(2x+5)(\frac{1}{2\sqrt{x}}) - \sqrt{x}(2)}{(2x+5)^2}\)

Step 2 ：Simplify the derivative: \(f'(x) = \frac{5 - 2\sqrt{x}}{4x^2 + 20x + 25}\)

Step 3 ：Find the slope of the tangent line at the point \((1, f(1))\) by plugging in \(x = 1\) into the derivative: \(m = \frac{5 - 2\sqrt{1}}{4*1^2 + 20*1 + 25} = -\frac{3}{49}\)

Step 4 ：Use the point-slope form of a line to find the equation of the tangent line: \(y - f(1) = -\frac{3}{49}(x - 1)\)

Step 5 ：\(\boxed{y - f(1) = -\frac{3}{49}(x - 1)}\)