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Câu hỏi: Cho hàm số $g\left( x \right)=\int\limits_{\sqrt{x}}^{{{x}^{2}}}{\sqrt{t}\sin tdt}$ xác định với mọi $x>0$. Tính $g'\left( x \right)$ được kết quả:
A. $g'\left( x \right)={{x}^{2}}\sin \left( {{x}^{2}} \right)-\dfrac{\sin \left( \sqrt{x} \right)}{\sqrt[4]{x}}.$
B. $g'\left( x \right)=2{{x}^{2}}\sin \left( {{x}^{2}} \right)-\dfrac{\sin \left( \sqrt{x} \right)}{2\sqrt[4]{x}}.$
C. $g'\left( x \right)=2{{x}^{2}}\sin \left( {{x}^{2}} \right)-\dfrac{\sin \left( \sqrt{x} \right)}{\sqrt[4]{x}}.$
D. $g'\left( x \right)={{x}^{2}}\sin \left( {{x}^{2}} \right)-\dfrac{\sin \left( \sqrt{x} \right)}{2\sqrt[4]{x}}.$
A. $g'\left( x \right)={{x}^{2}}\sin \left( {{x}^{2}} \right)-\dfrac{\sin \left( \sqrt{x} \right)}{\sqrt[4]{x}}.$
B. $g'\left( x \right)=2{{x}^{2}}\sin \left( {{x}^{2}} \right)-\dfrac{\sin \left( \sqrt{x} \right)}{2\sqrt[4]{x}}.$
C. $g'\left( x \right)=2{{x}^{2}}\sin \left( {{x}^{2}} \right)-\dfrac{\sin \left( \sqrt{x} \right)}{\sqrt[4]{x}}.$
D. $g'\left( x \right)={{x}^{2}}\sin \left( {{x}^{2}} \right)-\dfrac{\sin \left( \sqrt{x} \right)}{2\sqrt[4]{x}}.$
Ta gọi $F\left( t \right)$ là nguyên hàm của $\sqrt{t}\sin t$.
Ta có $g\left( x \right)=\int\limits_{\sqrt{x}}^{{{x}^{2}}}{\sqrt{t}\sin tdt}=F\left( {{x}^{2}} \right)-F\left( \sqrt{x} \right)$.
$\Rightarrow g'\left( x \right)=2{{x}^{2}}F'\left( {{x}^{2}} \right)-\dfrac{1}{2\sqrt{x}}F'\left( \sqrt{x} \right)\Rightarrow g'\left( x \right)=2{{x}^{2}}\sin \left( {{x}^{2}} \right)-\dfrac{\sin \left( \sqrt{x} \right)}{2\sqrt[4]{x}}$.
Ta có $g\left( x \right)=\int\limits_{\sqrt{x}}^{{{x}^{2}}}{\sqrt{t}\sin tdt}=F\left( {{x}^{2}} \right)-F\left( \sqrt{x} \right)$.
$\Rightarrow g'\left( x \right)=2{{x}^{2}}F'\left( {{x}^{2}} \right)-\dfrac{1}{2\sqrt{x}}F'\left( \sqrt{x} \right)\Rightarrow g'\left( x \right)=2{{x}^{2}}\sin \left( {{x}^{2}} \right)-\dfrac{\sin \left( \sqrt{x} \right)}{2\sqrt[4]{x}}$.
Đáp án B.