यदि f(x) = int_{9x^2}^{x^4} 5^{sqrt{t}} dt है | vedclass

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(B) Leibniz integral rule का उपयोग करते हुए,$f'(x) = \frac{d}{dx} \int_{g(x)}^{h(x)} F(t) dt = F(h(x)) \cdot h'(x) - F(g(x)) \cdot g'(x)$.यहाँ,$F(t) =

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$108(5^9)$

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(B) Leibniz integral rule का उपयोग करते हुए,$f'(x) = \frac{d}{dx} \int_{g(x)}^{h(x)} F(t) dt = F(h(x)) \cdot h'(x) - F(g(x)) \cdot g'(x)$.यहाँ,$F(t) = 5^{\sqrt{t}}$,$h(x) = x^4$,और $g(x) = 9x^2$.$f'(x) = 5^{\sqrt{x^4}} \cdot (4x^3) - 5^{\sqrt{9x^2}} \cdot (18x) = 5^{x^2} \cdot 4x^3 - 5^{3|x|} \cdot 18x$.$x = 3$ के लिए,$f'(3) = 5^{3^2} \cdot 4(3^3) - 5^{3(3)} \cdot 18(3) = 5^9 \cdot 108 - 5^9 \cdot 54 = 54 \cdot 5^9$.सीमा का मान $\lim_{h 	o 0} \frac{f(3 + h) - f(3 - h)}{h} = f'(3) - (-f'(3)) = 2f'(3)$ है।$2 \cdot (54 \cdot 5^9) = 108 \cdot 5^9$.

यदि $f(x) = \int_{9x^2}^{x^4} 5^{\sqrt{t}} dt$ है,तो $\lim_{h \to 0} \frac{f(3 + h) - f(3 - h)}{h}$ का मान ज्ञात कीजिए।

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