Use part one of the fundamental theorem of calculus to find the derivative of the function.
\[
g(x)=\int_{0}^{x} \sqrt{t^{3}+t^{5}} d t
\]
\[
g^{\prime}(x)=
\]
Answer
\(\boxed{g^{\prime}(x)=\sqrt{x^{5}+x^{3}}}\) is the final answer
Step 1 ：Given the function \(g(x)=\int_{0}^{x} \sqrt{t^{3}+t^{5}} d t\)
Step 2 ：We are asked to find the derivative of this function \(g(x)\)
Step 3 ：According to the Fundamental Theorem of Calculus, the derivative of this function \(g(x)\) is just the function inside the integral evaluated at x
Step 4 ：So, we need to replace t with x in the function inside the integral to get the derivative of \(g(x)\)
Step 5 ：Substituting \(t = x\) in the function inside the integral, we get \(g^{\prime}(x)=\sqrt{x^{5}+x^{3}}\)
Step 6 ：\(\boxed{g^{\prime}(x)=\sqrt{x^{5}+x^{3}}}\) is the final answer
