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Câu hỏi: Hàm số $f\left( x \right)={{\text{e}}^{\sqrt{{{x}^{2}}+1}}}$ có đạo hàm là
A. ${f}'\left( x \right)=\dfrac{2x}{\sqrt{{{x}^{2}}+1}}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}$.
B. ${f}'\left( x \right)=\dfrac{x}{\sqrt{{{x}^{2}}+1}}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}.\ln 2$.
C. ${f}'\left( x \right)=\dfrac{x}{2\sqrt{{{x}^{2}}+1}}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}$.
D. ${f}'\left( x \right)=\dfrac{x}{\sqrt{{{x}^{2}}+1}}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}$.
A. ${f}'\left( x \right)=\dfrac{2x}{\sqrt{{{x}^{2}}+1}}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}$.
B. ${f}'\left( x \right)=\dfrac{x}{\sqrt{{{x}^{2}}+1}}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}.\ln 2$.
C. ${f}'\left( x \right)=\dfrac{x}{2\sqrt{{{x}^{2}}+1}}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}$.
D. ${f}'\left( x \right)=\dfrac{x}{\sqrt{{{x}^{2}}+1}}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}$.
${f}'\left( x \right)={{\left( \sqrt{{{x}^{2}}+1} \right)}^{\prime }}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}=\dfrac{2x}{2\sqrt{{{x}^{2}}+1}}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}=\dfrac{x}{\sqrt{{{x}^{2}}+1}}.{{\text{e}}^{\sqrt{{{x}^{2}}+1}}}$.
Đáp án D.