Let S(x) = int_{x^2}^{x^3} ln t , dt for x | vedclass

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(B) Using the Leibniz integral rule,we differentiate $S(x)$ with respect to $x$: $S'(x) = \ln(x^3) \cdot \frac{d}{dx}(x^3) - \ln(x^2) \cdot \frac{d}{d

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derivable and continuous in its domain

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(B) Using the Leibniz integral rule,we differentiate $S(x)$ with respect to $x$: $S'(x) = \ln(x^3) \cdot \frac{d}{dx}(x^3) - \ln(x^2) \cdot \frac{d}{dx}(x^2)$ $S'(x) = (3 \ln x) \cdot (3x^2) - (2 \ln x) \cdot (2x)$ $S'(x) = 9x^2 \ln x - 4x \ln x$ $S'(x) = x \ln x (9x - 4)$ Now,$H(x) = \frac{S'(x)}{x} = \frac{x \ln x (9x - 4)}{x} = \ln x (9x - 4)$ Since $\ln x$ and $(9x - 4)$ are both continuous and differentiable for all $x > 0$,their product $H(x) = \ln x (9x - 4)$ is also continuous and differentiable in its domain $(0, \infty)$.

Let $S(x) = \int_{x^2}^{x^3} \ln t \, dt$ for $x > 0$ and $H(x) = \frac{S'(x)}{x}$. Then $H(x)$ is :

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