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Câu hỏi: Đạo hàm của hàm số $f\left( x \right)=\dfrac{{{3}^{x}}-1}{{{3}^{x}}+1}$ là
A. $f'\left( x \right)=-\dfrac{2}{{{\left( {{3}^{x}}+1 \right)}^{2}}}{{.3}^{x}}.$
B. $f'\left( x \right)=\dfrac{2}{{{\left( {{3}^{x}}+1 \right)}^{2}}}{{.3}^{x}}.$
C. $f'\left( x \right)=\dfrac{2}{{{\left( {{3}^{x}}+1 \right)}^{2}}}{{.3}^{x}}\ln 3.$
D. $f'\left( x \right)=-\dfrac{2}{{{\left( {{3}^{x}}+1 \right)}^{2}}}{{.3}^{x}}\ln 3.$
A. $f'\left( x \right)=-\dfrac{2}{{{\left( {{3}^{x}}+1 \right)}^{2}}}{{.3}^{x}}.$
B. $f'\left( x \right)=\dfrac{2}{{{\left( {{3}^{x}}+1 \right)}^{2}}}{{.3}^{x}}.$
C. $f'\left( x \right)=\dfrac{2}{{{\left( {{3}^{x}}+1 \right)}^{2}}}{{.3}^{x}}\ln 3.$
D. $f'\left( x \right)=-\dfrac{2}{{{\left( {{3}^{x}}+1 \right)}^{2}}}{{.3}^{x}}\ln 3.$
Ta có $f'\left( x \right)=\dfrac{{{3}^{x}}\ln 3\left( {{3}^{x}}+1 \right)-{{3}^{x}}\ln 3\left( {{3}^{x}}-1 \right)}{{{\left( {{3}^{x}}+1 \right)}^{2}}}=\dfrac{{{2.3}^{x}}\ln 3}{{{\left( {{3}^{x}}+1 \right)}^{2}}}$
Đáp án C.