Find $f$ such that $f^{\prime }(x)=\frac{8}{\sqrt{x}},f(16)=78$
$$f(x)=$$

Step 1 ：The problem is asking for a function $f(x)$ such that its derivative $f^{\prime }(x)$ is equal to $\frac{8}{\sqrt{x}}$ and $f(16)$ is equal to 78. This is a problem of finding the antiderivative (or integral) of a function and then using the given point to solve for the constant of integration.

Step 2 ：The antiderivative of $\frac{8}{\sqrt{x}}$ can be found using the power rule for integration, which states that the integral of $x^n$ is $\frac{1}{n+1}{x}^{n+1}$, where $n$ is any real number except -1. In this case, we can rewrite $\frac{8}{\sqrt{x}}$ as $8x^{-1/2}$, so the antiderivative will be $\frac{1}{-1/2+1}8x^{-1/2+1}$.

Step 3 ：After finding the antiderivative, we can use the given point $(16,78)$ to solve for the constant of integration. This is done by substituting the x and y values into the antiderivative equation and solving for the constant.

Step 4 ：The function $f(x)$ that satisfies the given conditions is $f(x)=16\sqrt{x}+14$.

Step 5 ：Final Answer: $\boxed{{\displaystyle f(x)=16\sqrt{x}+14}}$