If f(x) = intlimits_0^x {frac{1}{{sqrt {1 + { | vedclass

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(B) Given $f(x) = \int\limits_0^x \frac{1}{\sqrt{1+t^3}} dt$. By the Fundamental Theorem of Calculus,$f'(x) = \frac{1}{\sqrt{1+x^3}}$.Since $h(x)$ is

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$\frac{3}{2}$

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(B) Given $f(x) = \int\limits_0^x \frac{1}{\sqrt{1+t^3}} dt$. By the Fundamental Theorem of Calculus,$f'(x) = \frac{1}{\sqrt{1+x^3}}$.Since $h(x)$ is the inverse of $f(x)$,we have $f(h(x)) = x$.Differentiating both sides with respect to $x$,we get $f'(h(x)) \cdot h'(x) = 1$,which implies $h'(x) = \frac{1}{f'(h(x))} = \sqrt{1 + h^3(x)}$.Now,differentiate $h'(x) = (1 + h^3(x))^{1/2}$ with respect to $x$:$h''(x) = \frac{1}{2}(1 + h^3(x))^{-1/2} \cdot (3h^2(x) \cdot h'(x))$.Substitute $h'(x) = \sqrt{1 + h^3(x)}$ into the expression:$h''(x) = \frac{3h^2(x) \cdot \sqrt{1 + h^3(x)}}{2 \sqrt{1 + h^3(x)}} = \frac{3}{2} h^2(x)$.Therefore,$\frac{h''(x)}{h^2(x)} = \frac{3}{2}$.

If $f(x) = \int\limits_0^x {\frac{1}{{\sqrt {1 + {t^3}} }}\,} dt$ and $h(x)$ is the inverse of $f(x)$,then the value of $\frac{{h''(x)}}{{{h^2}(x)}}$ is

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