Let $g(x)=\int_{4}^{x}(5+\sqrt{t}) d t$. Find $g^{\prime}(x)$.
\[
g^{\prime}(x)=
\]

Answer

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Answer

Therefore, the final answer is $\boxed{g^{\prime}(x)=5+\sqrt{x}}$.

Steps

Step 1 ：Let $g(x)=\int_{4}^{x}(5+\sqrt{t}) d t$.

Step 2 ：By the Fundamental Theorem of Calculus, the derivative of the integral of a function from a constant to x is just the function evaluated at x.

Step 3 ：So, $g^{\prime}(x)=5+\sqrt{x}$.

Step 4 ：Therefore, the final answer is $\boxed{g^{\prime}(x)=5+\sqrt{x}}$.

Let $g(x)=\int_{4}^{x}(5+\sqrt{t}) d t$. Find $g^{\prime}(x)$.\[g^{\prime}(x)=\]

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Let $g(x)=\int_{4}^{x}(5+\sqrt{t}) d t$. Find $g^{\prime}(x)$.
\[
g^{\prime}(x)=
\]