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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 ${f}'\left( x \right)=\dfrac{{{3}^{x}}\ln 3\left( {{3}^{x}}+1 \right)-{{3}^{3}}\ln 3\left( {{3}^{x}}-1 \right)}{{{\left( {{3}^{x}}+1 \right)}^{2}}}=\dfrac{{{2.3}^{x}}\ln 3}{{{\left( {{3}^{x}}+1 \right)}^{2}}}.$
Đáp án C.