$\begin{array}{l}g(x)=\int_{1}^{x} \sqrt{2 t+7} d t \ g^{\prime}(9)=\end{array}$

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∻Solution by Steps∻ ‖ step 1 ⋮ Apply the Fundamental Theorem of Calculus Part 1 to differentiate the integral. ‖ ‖ step 2 ⋮ According to the theorem, the derivative of the integral from a constant to x of a function is the function evaluated at x. ‖ ‖ step 3 ⋮ So, $g'(x) = \frac{d}{dx} \int_{1}^{x} \sqrt{2t+7} \, dt = \sqrt{2x+7}$. ‖ ‖ step 4 ⋮ Evaluate the derivative at $x=9$: $g'(9) = \sqrt{2(9)+7}$. ‖ ‖ step 5 ⋮ Simplify the expression: $g'(9) = \sqrt{18+7} = \sqrt{25}$. ‖ ‖ step 6 ⋮ Calculate the final value: $g'(9) = 5$. ‖ ∻Answer∻ ⚹ $g'(9) = 5$ ⚹ ∻Key Concept∻ ⚹Fundamental Theorem of Calculus Part 1⚹ ∻Explanation∻ ⚹The derivative of the integral of a function from a constant to x is the function evaluated at x.⚹◊What is the derivative of the function $g(x)=\int_{1}^{x} \sqrt{2t+7} dt$ at $x=9$?⍭ Generate me a similar question◊

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