If f(x) = int_{9x^2}^{x^4} 5^{sqrt{t}} dt,the | vedclass

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(B) Using the Leibniz integral rule,$f'(x) = \frac{d}{dx} \int_{g(x)}^{h(x)} F(t) dt = F(h(x)) \cdot h'(x) - F(g(x)) \cdot g'(x)$.Here,$F(t) = 5^{\sqr

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$108(5^9)$

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(B) Using the Leibniz integral rule,$f'(x) = \frac{d}{dx} \int_{g(x)}^{h(x)} F(t) dt = F(h(x)) \cdot h'(x) - F(g(x)) \cdot g'(x)$.Here,$F(t) = 5^{\sqrt{t}}$,$h(x) = x^4$,and $g(x) = 9x^2$.$f'(x) = 5^{\sqrt{x^4}} \cdot (4x^3) - 5^{\sqrt{9x^2}} \cdot (18x) = 5^{x^2} \cdot 4x^3 - 5^{3|x|} \cdot 18x$.For $x = 3$,$f'(3) = 5^{3^2} \cdot 4(3^3) - 5^{3(3)} \cdot 18(3) = 5^9 \cdot 108 - 5^9 \cdot 54 = 54 \cdot 5^9$.The limit is $\lim_{h 	o 0} \frac{f(3 + h) - f(3 - h)}{h} = f'(3) - (-f'(3)) = 2f'(3)$.$2 \cdot (54 \cdot 5^9) = 108 \cdot 5^9$.

If $f(x) = \int_{9x^2}^{x^4} 5^{\sqrt{t}} dt$,then $\lim_{h \to 0} \frac{f(3 + h) - f(3 - h)}{h}$ is equal to

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