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