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

Answer

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Answer

$\boxed{y - f(1) = -\frac{3}{49}(x - 1)}$

Steps

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

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

Answer

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