The derivative of F(x) = int_{x^2}^{x^3} frac | vedclass

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(D) We use the Leibniz Integral Rule: $\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) \, dt = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$.Given $F(x) = \int_{x^2

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$(\log x)^{-1} \cdot x(x - 1)$

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(D) We use the Leibniz Integral Rule: $\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) \, dt = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$.Given $F(x) = \int_{x^2}^{x^3} \frac{1}{\log t} \, dt$,we have $f(t) = \frac{1}{\log t}$,$h(x) = x^3$,and $g(x) = x^2$.Then $h'(x) = 3x^2$ and $g'(x) = 2x$.Applying the rule:$F'(x) = \frac{1}{\log(x^3)} \cdot (3x^2) - \frac{1}{\log(x^2)} \cdot (2x)$Since $\log(x^n) = n \log x$,we get:$F'(x) = \frac{3x^2}{3 \log x} - \frac{2x}{2 \log x}$$F'(x) = \frac{x^2}{\log x} - \frac{x}{\log x} = \frac{x^2 - x}{\log x} = \frac{x(x - 1)}{\log x} = x(x - 1)(\log x)^{-1}$.

The derivative of $F(x) = \int_{x^2}^{x^3} \frac{1}{\log t} \, dt$,$(x > 0)$ is

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