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F(x)=int_(2)^(2x)sqrt(15-t)dt F^(')(x)=

F(x)=∫22x15−tdt F(x)=\int_{2}^{2 x} \sqrt{15-t} d t \newlineF′(x)= F^{\prime}(x)=

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Evaluate definite integrals using the power ruleright chevron icon

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Q. F(x)=∫22x15−tdt F(x)=\int_{2}^{2 x} \sqrt{15-t} d t \newlineF′(x)= F^{\prime}(x)=
  1. Apply Fundamental Theorem of Calculus: Use the Fundamental Theorem of Calculus Part 11, which states that if F(x)=∫ag(x)f(t) dtF(x) = \int_{a}^{g(x)}f(t)\,dt, then F′(x)=f(g(x))⋅g′(x)F'(x) = f(g(x)) \cdot g'(x).
  2. Identify f(t)f(t) and g(x)g(x): Identify f(t)f(t) as 15−t\sqrt{15-t} and g(x)g(x) as 2x2x.
  3. Calculate g′(x)g'(x): Calculate g′(x)g'(x), which is the derivative of 2x2x with respect to xx.\newlineg′(x)=ddx(2x)=2g'(x) = \frac{d}{dx}(2x) = 2.
  4. Substitute into F′(x)F'(x): Substitute g(x)=2xg(x) = 2x and g′(x)=2g'(x) = 2 into the formula F′(x)=f(g(x))⋅g′(x)F'(x) = f(g(x)) \cdot g'(x).\newlineF′(x)=15−2x⋅2F'(x) = \sqrt{15 - 2x} \cdot 2.
  5. Simplify F′(x)F'(x): Simplify the expression for F′(x)F'(x).F′(x)=2⋅15−2xF'(x) = 2 \cdot \sqrt{15 - 2x}.

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