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Câu hỏi: Đạo hàm của hàm số $f\left( x \right)=\dfrac{{{2}^{x}}-1}{{{2}^{x}}+1}$ là
A. $\dfrac{{{2}^{x+1}}\ln 2}{{{\left( {{2}^{x}}+1 \right)}^{2}}}$.
B. $\dfrac{{{2}^{x}}\ln 2}{{{\left( {{2}^{x}}+1 \right)}^{2}}}$.
C. $\dfrac{{{2}^{x+1}}}{{{\left( {{2}^{x}}+1 \right)}^{2}}}$.
D. $\dfrac{{{2}^{x}}}{{{\left( {{2}^{x}}+1 \right)}^{2}}}$.
A. $\dfrac{{{2}^{x+1}}\ln 2}{{{\left( {{2}^{x}}+1 \right)}^{2}}}$.
B. $\dfrac{{{2}^{x}}\ln 2}{{{\left( {{2}^{x}}+1 \right)}^{2}}}$.
C. $\dfrac{{{2}^{x+1}}}{{{\left( {{2}^{x}}+1 \right)}^{2}}}$.
D. $\dfrac{{{2}^{x}}}{{{\left( {{2}^{x}}+1 \right)}^{2}}}$.
Ta có ${f}'\left( x \right)=\dfrac{{{\left( {{2}^{x}}-1 \right)}^{\prime }}\left( {{2}^{x}}+1 \right)-\left( {{2}^{x}}-1 \right){{\left( {{2}^{x}}+1 \right)}^{\prime }}}{{{\left( {{2}^{x}}+1 \right)}^{2}}}=\dfrac{{{2}^{x}}\ln 2\left( {{2}^{x}}+1 \right)-\left( {{2}^{x}}-1 \right){{2}^{x}}\ln 2}{{{\left( {{2}^{x}}+1 \right)}^{2}}}$ $=\dfrac{{{2}^{x+1}}\ln 2}{{{\left( {{2}^{x}}+1 \right)}^{2}}}$
Đáp án A.