Suppose $f(x)=\sqrt{4 x^{2}+5 x+3}$. Find:
\[
f^{\prime}(5)=
\]

Step 1 ：Identify the outer function as \(f(u)=\sqrt{u}\) and the inner function as \(g(x)=4 x^{2}+5 x+3\)

Step 2 ：Find the derivative of the outer function as \(f'(u)=\frac{1}{2\sqrt{u}}\)

Step 3 ：Find the derivative of the inner function as \(g'(x)=8x+5\)

Step 4 ：Apply the chain rule to find the derivative of the function as \(f'(x)=\frac{1}{2\sqrt{4 x^{2}+5 x+3}} \cdot (8x+5)\)

Step 5 ：Substitute \(x=5\) into \(f'(x)\) to find \(f'(5)\) as \(f'(5)=\frac{1}{2\sqrt{4 \cdot 5^{2}+5 \cdot 5+3}} \cdot (8 \cdot 5+5)\)

Step 6 ：Simplify the expression under the square root to get \(f'(5)=\frac{1}{2\sqrt{100+25+3}} \cdot (40+5)\)

Step 7 ：Further simplify to get \(f'(5)=\frac{1}{2\sqrt{128}} \cdot 45\)

Step 8 ：Simplify the denominator to get \(f'(5)=\frac{45}{2 \cdot 8 \sqrt{2}}\)

Step 9 ：Further simplify to get \(f'(5)=\frac{45}{16\sqrt{2}}\)

Step 10 ：Rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{2}\) to get \(f'(5)=\frac{45\sqrt{2}}{16 \cdot 2}\)

Step 11 ：Finally, simplify to get \(f'(5)=\frac{45\sqrt{2}}{32}\)

Step 12 ：\(\boxed{f'(5)=\frac{45\sqrt{2}}{32}}\)