Use the formula $$ f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x} $$ to find the derivative of the function. $$ g(x)=5 x+\sqrt{x} $$

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We start with the definition of the derivative:
$$
f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x}.
$$
Here, the function is 
$$
g(x)=5x+\sqrt{x}.
$$
Thus, we have
$$
g^{\prime}(x)=\lim_{z\rightarrow x}\frac{g(z)-g(x)}{z-x}=\lim_{z\rightarrow x}\frac{\left(5z+\sqrt{z}\right)-\left(5x+\sqrt{x}\right)}{z-x}.
$$

We can split the fraction into two parts:
$$
g^{\prime}(x)=\lim_{z\rightarrow x}\left[\frac{5z-5x}{z-x}+\frac{\sqrt{z}-\sqrt{x}}{z-x}\right].
$$

1. For the first term:
   $$
   \frac{5z-5x}{z-x}=5\frac{z-x}{z-x}=5.
   $$
   Hence, 
   $$
   \lim_{z\rightarrow x} \frac{5z-5x}{z-x}=5.
   $$

2. For the second term:
   $$
   \lim_{z\rightarrow x}\frac{\sqrt{z}-\sqrt{x}}{z-x}.
   $$
   Multiply the numerator and the denominator by the conjugate $$\sqrt{z}+\sqrt{x}$$:
   $$
   \frac{\sqrt{z}-\sqrt{x}}{z-x} \cdot \frac{\sqrt{z}+\sqrt{x}}{\sqrt{z}+\sqrt{x}}=\frac{z-x}{(z-x)(\sqrt{z}+\sqrt{x})}=\frac{1}{\sqrt{z}+\sqrt{x}}.
   $$
   Now taking the limit as $$z \rightarrow x$$:
   $$
   \lim_{z\rightarrow x}\frac{1}{\sqrt{z}+\sqrt{x}}=\frac{1}{2\sqrt{x}}.
   $$

Combining both parts, we obtain:
$$
g^{\prime}(x)=5+\frac{1}{2\sqrt{x}}.
$$
The derivative of $$ g(x) = 5x + \sqrt{x} $$ is $$ g^{\prime}(x) = 5 + \frac{1}{2\sqrt{x}} $$.

Use the formula to find the derivative of the function.

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