Answer
$$
f^{\prime}(x)=\sqrt{x}(6+5 x), \quad f(1)=10
$$
The required function is
$$
f(x) =4 x^{3 / 2}+2 x^{5 / 2}+4
$$
Work Step by Step
$$
f^{\prime}(x)=\sqrt{x}(6+5 x), \quad f(1)=10
$$
can be written as the following:
$$
f^{\prime}(x)=\sqrt{x}(6+5 x)=6 x^{1 / 2}+5 x^{3 / 2}
$$
The general anti-derivative of $
f^{\prime }(x)=\sqrt{x}(6+5 x)
$ is
$$
f(x)=4 x^{3 / 2}+2 x^{5 / 2}+C
$$
To determine C we use the fact that $f(1)=10$:
$$
f(1)=4 (1)^{3 / 2}+2 (1)^{5 / 2}+C =10
$$
$ \Rightarrow $
$$
6+C=10 \quad \Rightarrow \quad C=4,
$$
so
$$
f(x) =4 x^{3 / 2}+2 x^{5 / 2}+4
$$
